# How to Prove It - Solutions

## Chapter - 6, Mathematical Induction

### Summary

Note: In this book natural numbers include $0$ also. But it seems like in some places in the proofs, I messed up with this convention. In some placed I might have considered $N$ includes $0$ while in other places vice versa. Please point out to me, I will correct to use the books version of $\mathbb N$ in all the places.

• Recursive functions are defined in the following way:
• For some values, x, f(x) is given.
• For values other then given, f(x) is defined by calling itself for lower/upper values.
• Thus by recursively tracing down he values we reach to a value for which f(x) is given and then move back filling in the corresponding values.
• This section shows that recursion and induction are similar by using some examples.

Soln1

Lets try by putting some values:

$x$ $f(x)$
$1$ $\frac 1 2$
$2$ $\frac 2 3$
$3$ $\frac 3 4$
$4$ $\frac 4 5$
$5$ $\frac 5 6$
$6$ $\frac 6 7$

Thus it appears that $f(x) = \frac x {x+1}$.

Lets prove this by mathematical induction.

Base Case:

It is directly clear from above table.

Induction step:

Suppose it is true for $n$. Thus $\sum_{i=1}^n {\frac 1 {i(i+1)}} = \frac n {n+1}$.

Now for $n + 1$:

We have $\sum_{i=1}^{n+1} {\frac 1 {i(i+1)}} = \sum_{i=1}^n {\frac 1 {i(i+1)}} + \frac {1} {(n+1)(n+1 + 1)}$. Using induction hypothesis $\sum_{i=1}^n {\frac 1 {i(i+1)}} = \frac n {n+1}$, we get:

$= \frac n {n+1} + \frac {1} {(n+1)(n+2)}$.
$= \frac {n(n+2) + 1} {(n+1)(n+2)}$.
$= \frac {n^2 + 2n + 1} {(n+1)(n+2)}$.
$= \frac {(n+1)(n+1)} {(n+1)(n+2)}$.
$= \frac {(n+1)} {(n+2)}$.
$= \frac {(n+1)} {(n+1 + 1)}$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true. Thus formulae is correct.

Soln2

Base Case:

For $n = 1$, we have $\sum_{i=1}^{n} {\frac 1 {i(i+1)(i+2)}} = \frac 1 {1(1+1)(1+2)} = \frac 1 6 = \frac {1^2 + 3 \cdot 1} {4(1+1)(1+2)}$. Thus $P(1)$ is true.

Induction Step:

Suppose theorem is correct for $n$. Thus $\sum_{i=1}^{n} {\frac 1 {i(i+1)(i+2)}} = \frac {n^2 + 3n} {4(n+1)(n+2)}$.

Now for $n+1$, we have:

$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \sum_{i=1}^{n} {\frac 1 {i(i+1)(i+2)}} + \frac 1 {(n+1)(n+2)(n+3)}$.
Using induction hypothesis:
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac {n^2 + 3n} {4(n+1)(n+2)} + \frac 1 {(n+1)(n+2)(n+3)}$.
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac {(n^2+3n)(n+3) + 4} {4(n+1)(n+2)(n+3)}$.
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac {n^3 + 3n^2 + 3n^2 + 9n + 4} {4(n+1)(n+2)(n+3)}$.
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac {(n+1)(n^2+5n+4)} {4(n+1)(n+2)(n+3)}$.
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac {(n+1)(n+1)(n+4)} {4(n+1)(n+2)(n+3)}$.
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac {(n+1)(n+4)} {4(n+2)(n+3)}$.
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac {(n+1)(n+1+3)} {4(n+2)(n+3)}$.
$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i(i+1)(i+2)}} = \frac { {(n+1)}^2 + 3(n+1)} {4(n+1+1)(n+1+2)}$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln3

Base Case:

For $n = 2$, we have $\sum_{i=2}^{n} {\frac 1 {(i-1)(i+1)}} = \frac 1 {(2-1)(2+1)} = \frac 1 3 = \frac {3 \cdot 2^2 - 2 - 2} {4 \cdot 2 (2+1)}$.

Thus $P(2)$ is true.

Induction step:

Suppose theorem is correct for $n$. Thus $\sum_{i=2}^{n} {\frac 1 {(i-1)(i+1)}} = \frac {3 n^2 - n - 2} {4 n (n+1)}$.

Now for $n+1$, we have:

$\Rightarrow \sum_{i=2}^{n+1} {\frac 1 {(i-1)(i+1)}} = {( \sum_{i=2}^{n} {\frac 1 {(i-1)(i+1)}} )} + \frac 1 {(n+1-1)(n+1+1)}$.
Using induction hypothesis:
$= \frac {3 n^2 - n - 2} {4 n (n+1)} + \frac 1 {(n+1-1)(n+1+1)}$.
$= \frac {3 n^2 - n - 2} {4 n (n+1)} + \frac 1 {(n)(n+2)}$.
$= \frac {3n^3 + 5n^2} {4n(n+1)(n+2)}$.
$= \frac {n^2(3n + 5)} {4n(n+1)(n+2)}$.
$= \frac {n(3n + 5)} {4(n+1)(n+2)}$.
$= \frac {(n+1-1)(3(n+1-1) + 5)} {4(n+1)(n+2)}$.
$= \frac {(n+1-1)(3(n+1) + 2)} {4(n+1)(n+2)}$.
$= \frac { {(n+1)}^2 - 3(n+1) - 2} {4(n+1)(n+1+1)}$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln4

Base Case:

For $n = 0$, we have $\sum_{i=0}^{n} {(2i+1)}^2 = 1 = \frac {(0+1)(2 \times 0 + 1)(2 \times 0 + 3)} 3$.

Thus $P(0)$ is true.

Induction Step:

Suppose theorem is correct for $n$. Thus $\sum_{i=0}^{n} {(2i+1)}^2 = \frac {(n+1)(2n + 1)(2n + 3)} 3$.

Now for $n+1$, we have:
$\Rightarrow \sum_{i=0}^{n+1} {(2i+1)}^2 = {( \sum_{i=0}^{n} {(2i+1)}^2 )} + {(2(n+1)+1)}^2$.
$= {\frac {(n+1)(2n + 1)(2n + 3)} 3} + {(2n + 3)}^2$.
$= \frac {(n+1)(2n + 1)(2n + 3) + 3{(2n+3)}^2} 3$.
$= \frac {(2n + 3)((n+1)(2n + 1) + 3(2n+3))} 3$.
$= \frac {(2n + 3)(2n^2 + n + 2n + 1 + 6n + 9)} 3$.
$= \frac {(2n + 3)(2n^2 + 9n + 10)} 3$.
$= \frac {(2n + 3)(2n+5)(n+2)} 3$.
$= \frac {(n+1+1)(2(n+1)+1)(2(n+1)+3)} 3$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln5

Suppose $r \ne 1$ is an arbitrary real number.

Base Case:

For $n = 0$, we have $\sum_{i = 0}^{n} r^i = r^0 = 1 = \frac {r^{0+1} - 1} {r-1}$.

Induction Step:

Suppose theorem is correct for $n$. Thus $\sum_{i = 0}^{n} r^i = \frac {r^{n+1} - 1} {r-1}$.

Now for $n+1$, we have:
$\Rightarrow \sum_{i = 0}^{n+1} r^i = \sum_{i = 0}^{n} r^i + r^{n+1}$.
$= { \frac {r^{n+1} - 1} {r-1} } + r^{n+1}$.
$= \frac {(r^{n+1} - 1) + r^{n+1}(r-1) } {r-1}$.
$= \frac {r^{n+1} - 1 + r^{n+2} - r^{n+1} } {r-1}$.
$= \frac { r^{n+2} - 1 } {r-1}$.
$= \frac { r^{n+1+1} - 1 } {r-1}$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln6

Base Case:

For $n = 1$, we have $\sum_{i=1}^{n} {\frac 1 {i^2}} = \frac 1 1^2 = 1$. Also $2 - \frac 1 1 = 1$.
Thus $\sum_{i=1}^{n} {\frac 1 {i^2}} \le$ 2 - \frac 1 n $$. Induction Step: Suppose theorem is correct for $n$. Thus $\sum_{i=1}^{n} {\frac 1 {i^2}} \le$ 2 - \frac 1 n$$.

Now for $n + 1$, we have:

$\Rightarrow \sum_{i=1}^{n+1} {\frac 1 {i^2}} = \sum_{i=1}^{n} {\frac 1 {i^2}} + { \frac 1 { {(n+1)}^2 } }$.
$\le { 2 - \frac 1 n } + { \frac 1 { {(n+1)}^2 } }$.
$= 2 - \frac {(n+1)^2 - n} { n{(n+1)}^2 }$.
$= 2 - \frac {n^2 + n + 1} { n{(n+1)}^2 }$.
$\le 2 - \frac {n^2 + n} { n{(n+1)}^2 }$.
$= 2 - \frac 1 { n+1 }$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln7

(a)

Base Case:

For $n = 0$, we have $\sum_{i=0}^{n} (a_i + b_i) = a_0 + b_0 = { \sum_{i=0}^{n} a_i } + { \sum_{i=0}^{n} b_i }$.

Induction Step:

Suppose it is true for $n$. Thus $\sum_{i=0}^{n} (a_i + b_i) \; = \; { \sum_{i=0}^{n} a_i } + { \sum_{i=0}^{n} b_i }$.

For $n+1$, we have:

$\Rightarrow \sum_{i=0}^{n+1} (a_i + b_i) = (\sum_{i=0}^{n} (a_i + b_i)) + (a_{n+1} + b_{n+1})$.
$= { \sum_{i=0}^{n} a_i } + { \sum_{i=0}^{n} b_i } + a_{n+1} + b_{n+1}$.
$= { \sum_{i=0}^{n} a_i } + a_{n+1} + { \sum_{i=0}^{n} b_i } + b_{n+1}$.
$= { \sum_{i=0}^{n+1} a_i } + { \sum_{i=0}^{n+1} b_i }$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

(b)

Base Case:

For $n = 0$, we have $c \cdot { \sum_{i=0}^{n} a_i } = c \cdot a_0 = \sum_{i=0}^{n} { c \cdot a_i }$.

Induction Step:

Suppose it is true for $n$. Thus $c \cdot { \sum_{i=0}^{n} a_i } = \sum_{i=0}^{n} { c \cdot a_i }$.

For $n + 1$, we have:

$c \cdot { \sum_{i=0}^{n+1} a_i } = c \cdot { \sum_{i=0}^{n} a_i } + c \cdot a_{n+1}$.
$= \sum_{i=0}^{n} { c \cdot a_i } + c \cdot a_{n+1}$.
$= \sum_{i=0}^{n+1} { c \cdot a_i }$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln8

(a)

Suppose $m$ is arbitrary natural number.

Base Case:

For $n = m$, we have ${\sum_{i=0}^{n} \frac 1 i} - {\sum_{i=0}^{m} \frac 1 i} = 0 = \frac {n-m} n$.

Induction Step:

Suppose theorem is correct for $n \ge m$. Thus ${\sum_{i=0}^{n} \frac 1 i} - {\sum_{i=0}^{m} \frac 1 i} = \frac {n-m} n$.

For $n + 1$, we have:

$\Rightarrow {\sum_{i=0}^{n+1} \frac 1 i} - {\sum_{i=0}^{m} \frac 1 i} = {\sum_{i=0}^{n} \frac 1 i} + { \frac 1 {n+1} } - {\sum_{i=0}^{m} \frac 1 i}$.
$= {\sum_{i=0}^{n} \frac 1 i} - {\sum_{i=0}^{m} \frac 1 i} + { \frac 1 {n+1} }$.
$\ge {\frac {n-m} n } + { \frac 1 {n+1} }$.
$= \frac { (n-m)(n+1) + n } {n(n+1)}$.
$= \frac {n^2 + n - mn - m + n} {n(n+1)}$.
$= \frac {n^2 + (n - m) - mn + n} {n(n+1)}$.
Since $n \ge m$,
$\ge \frac {n^2 - mn + n} {n(n+1)}$.
$\ge \frac {n - m + 1} {n+1}$.
$\ge \frac {n + 1 - m} {n+1}$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

(b)

Base Case:

For $n = 0$, we have $H_{2^n} = H_1 = 1 \ge 1 = 1 + n/2$.

Induction step:

Suppose theorem is correct for $n$. Thus $H_{2^n} \ge 1 + n/2$.

Since $2^{n+1} > 2^n$. Thus applying part(a), we have:

$H_{2^{n+1} } - H_{2^n} \ge \frac {2^{n+1} - 2^n } { 2^{n+1} } = \frac 1 2$.

Thus $H_{2^{n+1} } \ge H_{2^n} + { \frac 1 2 }$.
By induction hypothesis,
$\Rightarrow H_{2^{n+1} } \ge 1 + {\frac n 2 }+ { \frac 1 2 }$. $= 1 + { \frac {n+1} 2 }$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

(c)

We know that $\lim_{n \to \infty } {(1 + { \frac n 2 } ) } = \infty$.

Thus using part(b) $\lim_{n \to \infty} H_{2^n} = \infty$. Now from part(a) we can see that $H_n$ is a increasing sequence. Thus $\lim_{n \to \infty} H_{n} = \infty$

Soln9

Base Case:

For $n = 2$, we have $nH_n - n = 2( { \frac 1 1 } + { \frac 1 2 } ) - 2 = 3 - 2 = 1 = H_1 = \sum_{k=1}^{n-1} H_k$.

Induction Step:

Suppose theorem is true for $n$. Thus $\sum_{k=1}^{n-1} H_k = nH_n - n$.

Now for $n + 1$, we have $\sum_{k=1}^{n+1-1} H_k = \sum_{k=1}^{n} H_k$.
$= \sum_{k=1}^{n-1} H_k + H_n$.
$= nH_n - n + H_n$.
$= (n+1)H_n - n$.
$= ((n+1) \cdot \sum_{i=0}^{n} {\frac 1 i}) - n$.
$= ((n+1) \cdot \sum_{i=0}^{n} {\frac 1 i}) + 1 - 1 - n$.
$= ((n+1) \cdot \sum_{i=0}^{n} {\frac 1 i}) + (n+1){ \frac 1 {n+1} } - 1 - n$.
$= (n+1) \cdot (\sum_{i=0}^{n} {\frac 1 i} + { \frac 1 {n+1} }) - 1 - n$.
$= (n+1) \cdot (\sum_{i=0}^{n+1} {\frac 1 i} ) - 1 - n$.
$= (n+1) \cdot (\sum_{i=0}^{n+1} {\frac 1 i} ) - (n + 1)$.
$= (n+1)H_{n+1} - (n + 1)$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln10

Lets try by putting some values:

$n$ $\sum_{i=1}^n (i \cdot i!)$
$1$ $1 = 2! - 1$
$2$ $5 = 3! - 1$
$3$ $23 = 4! - 1$
$4$ $119 = 5! - 1$
$5$ $719 = 6! - 1$
$6$ $5039 = 7! - 1$

Thus it appears that $\sum_{i=1}^n (i \cdot i!) = (n+1)! - 1$.

Base Case:

It directly follows from above table.

Induction Step:

Suppose it is correct for $n$. Thus $\sum_{i=1}^n (i \cdot i!) = (n+1)! - 1$.

For $n+1$, we have:

$\sum_{i=1}^{n+1} (i \cdot i!) = \sum_{i=1}^n (i \cdot i!) + (n+1)\cdot(n+1)!$.
$= (n+1)! - 1 + (n+1)\cdot(n+1)!$.
$= (n+1+1)\cdot(n+1)! - 1$.

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln11

Lets try by putting some values:

$n$ $\sum_{i=1}^n {\frac i {(i+1)!} }$
$0$ $\frac 0 1 = \frac {1!-1} {1!}$
$1$ $\frac 1 2 = \frac {2!-1} {2!}$
$2$ $\frac 5 6 = \frac {3!-1} {3!}$
$3$ $\frac {23} {24} = \frac {4!-1} {4!}$
$4$ $\frac {119} {120} = \frac {5!-1} {5!}$
$5$ $\frac {719} {720} = \frac {6!-1} {6!}$

Thus it appears that $\sum_{i=1}^n {\frac i {(i+1)!} } = \frac {(n+1)!-1} {(n+1)!}$.

Base Case:

It directly follows from above table.

Induction Step:

Suppose it is correct for $n$. Thus $\sum_{i=1}^n {\frac i {(i+1)!} } = \frac {(n+1)!-1} {(n+1)!}$.

For $n+1$, we have:

$\sum_{i=1}^{n+1} {\frac i {(i+1)!} } = \sum_{i=1}^n {\frac i {(i+1)!} } + { \frac {n+1} {(n+1+1)!} }$.
$= { \frac {(n+1)!-1} {(n+1)!} } + { \frac {n+1} {(n+1+1)!} }$
$= { \frac {(n+1)!-1} {(n+1)!} } + { \frac {n+1} {(n+2)(n+1)!} }$
$= \frac {(n+2)((n+1)!-1) + (n+1)} {(n+2)(n+1)!}$
$= \frac {(n+2)(n+1)! - (n+2) + (n+1)} {(n+2)(n+1)!}$
$= \frac {(n+2)! - 1} {(n+2)!}$
$= \frac {(n+1+1)! - 1} {(n+1+1)!}$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln12

(a)

If we prove that $2^n \ge n+1$, then it is easily followed that $2^n \ge n$.

Base Case:

For $n = 0$, we have $2^n = 1 \ge (n+1)$. Thus $P(0)$ is true.

Inductive step:

Suppose theorem is correct for $n$. Thus $2^n \ge n + 1$.

For $n+1$, we have $2^{n+1} = 2 \cdot 2^n \ge 2(n+1) = 2n + 2 \ge n + 2 = n + 1 + 1$.

Thus we can conclude $2^n \ge n + 1$. It follows $2^n \ge n$.

(b)

Base Case:

For $n = 9$. We have $n! = 362880 ≥ 262144 = { (2^n) }^2$.

Induction Step:

Suppose theorem is correct for $n$. Thus $n! \ge { (2^n) }^2$.

For $n+1$, we have:

$(n+1)! = (n+1)n!$
$\ge (9+1)n!$
By induction hypothesis,
$\ge 10 \cdot { (2^n) }^2$
$\ge 4 \cdot { (2^n) }^2$
$= 2^2 \cdot { (2^n) }^2$
$= 2^{2n+2}$
$= { 2^{n+1} }^2$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

(c)

Base Case:

For $n = 0$, we have $0! = 1 \le 1 = 2^{0^2}$.

Induction step:

Suppose theorem is correct for $n$. Thus $n! \le 2^{n^2}$.

Now consider for $n+1$, we jave:

$(n+1)! = (n+1)n! \le (n+1) \cdot 2^{n^2}$.
From part(a), $n+1 \le 2^n$, we have:

$(n+1)! \le 2^n \cdot 2^{n^2}$.
$= 2^{ n^2 + n }$.
$\le 2^{ n^2 + n + n + 1}$.
$= 2^{n^2+2n+1}$.
$= 2^{ { (n+1) }^2}$.

Soln13

(a)

Base Case:

For $n = 0$, we have $(k^2 + n)! = (k^2)! \ge 1 = k^{2n}$.

Induction Step:

Suppose theorem is correct. Thus $(k^2 + n)! \ge k^{2n}$.

For $n+1$, we have $(k^2 + n + 1)!$:
$= (k^2+n+1)(k^2+n)!$
$\ge (k^2+n+1)k^{2n}$
$\ge (k^2)k^{2n}$
$= k^{2n+2}$
$= k^{2(n+1)}$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

(b)

Base Case:

Putting $n = k^2$ in part(a), we get $(2k^2)! \ge k^{2k^2}$. Thus for the given theorem if we assume $n = 2k^2$, then is follows $P(2k^2)$ is correct.

Induction Step:

Suppose theorem is correct. Thus $n! \ge k^n$.

Now consider $(n+1)!$:

$= (n+1)n!$
$\ge (n+1)k^n$
Since $n \ge 2k^2$,
$\ge (2k^2+1)k^n$
Since $k$ is positive integer,
$\ge (k)k^n$
$= k^{n+1}$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln14

Suppose $a$ is an arbitrary real number. Suppose $m$ is arbitrary integer.

Base Case:

For $n = 1$, we have ${(a^{m})}^n = a^m = a^{mn}$.

Induction step:

Suppose theorem is correct for $n$. Thus ${(a^m)}^n = a^{mn}$.

Thus ${(a^m)}^{n+1} = {(a^m)}^n \cdot a^m$
$a^{mn} \cdot a^m$
$a^{mn+m}$
$a^{m(n+1)}$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln15

Base Case:

For $n = 0$, $2^0 - 0 - 1 = 0 = a_0$.

Induction Step:

Suppose theorem is correct for $n$. Thus $a_n = 2^n - n - 1$.

Now, $a_{n+1} = 2a_n + n$
$= 2(2^n - n - 1) + n$
$= 2^{n+1} - 2n - 2 + n$
$= 2^{n+1} - n - 2$
$= 2^{n+1} - n - 1 - 1$
$= 2^{n+1} - (n + 1) - 1$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln16

Lets try by putting some values:

$n$ $a_n$
$0$ $2 = 2^{2^0}$
$1$ $4 = 2^{2^1}$
$2$ $16 = 2^{2^2}$
$3$ $256 = 2^{2^3}$
$4$ $65536 = 2^{2^4}$
$5$ $4294967296 = 2^{2^5}$

Thus it appears that $a_n = 2^{2^n}$.

Base Case:

It directly follows from table.

Induction step:

Suppose theorem is correct for $n$. Thus $a_n = 2^{2^n}$.

Now $a_{n+1} = {(a_n)}^2$
$= {(2^{2^n})}^2$
$= {(2^{2^n})}^2$
$= (2^{2 \cdot 2^n})$
$= (2^{2^{n+1}})$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln17

$n$ $a_n$
$1$ $1$
$2$ $\frac 1 2$
$3$ $\frac 1 3$
$4$ $\frac 1 4$
$5$ $\frac 1 5$

Thus it appears that $a_n = \frac 1 n$.

Base Case:

It directly follows from table.

Induction step:

Suppose theorem is correct for $n$. Thus $a_n = \frac 1 n$.

Now $a_{n+1} = \frac {a_n} {a_n + 1}$
$= \frac {\frac 1 n} { {\frac 1 n} + 1}$
$= \frac {\frac 1 n} { \frac {n+1} n }$
$= {\frac 1 n} \times { \frac n {n+1} }$
$= \frac 1 {n+1}$

Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln18

(a)

Expanding $\binom{n} {0} = \frac {n!} {0!(n-0)!} = 1$.

Also $\binom{n} {n} = \frac {n!} {n!(n-n)!} = 1$.

Thus $\binom{n} {0} = \binom{n} {n}$.

(b)

Lets compute ${\binom {n} {k} } + {\binom {n} {k-1} }$
$= {\frac {n!} {k!(n-k)!} } + {\frac {n!} {(k-1)!(n-k+1)!} }$
$= {\frac {n!} {k!(n-k)!} } + {\frac {n!} {(k-1)!(n-k+1)!} }$
$= {\frac {n!} {k(k-1)!(n-k)!} } + {\frac {n!} {(k-1)!(n-k+1)(n-k)!} }$
$= n! \times \frac {(n-k+1) + k} {k(n-k+1)(k-1)!(n-k)!}$
$= n! \times \frac {n+1} {k(n-k+1)(k-1)!(n-k)!}$
$= n! \times \frac {n+1} {k!(n-k+1)!}$
$= \frac {(n+1)!} {k!(n-k+1)!}$
$= \binom {(n+1)} {k}$

Hence proved.

(c)

Base Case:

For $n = 0$. Suppose $A$ has a set with $0$ elements. Thus only possible value of $k = 0$. Thus number of subsets of $A$ containing $0$ elements are $1$(empty set). Also $\binom n k = \binom 0 0 = 1$.

Induction Step:

Suppose theorem is correct if $A$ has $n$ elements. Thus number of $k$ sized subsets from $A$ are $\binom n k$.

Now suppose $A$ is a set of $n + 1$ elements. Suppose $a$ is an arbitrary element of $A$. Suppose $A' = A \setminus \{a\}$. Suppose $k$ is an integer such that $n+1 \ge k \ge 0$. We have following possible cases:

• Case 1: $k = 0$
Clearly there is only one possible zero sized subset of $A$. Also, $\binom {n+1} 0 = 1$. Thus for $k = 0$, number of subsets of $A$ is same as $\binom {n+1} k$.

• Case 2: $k = n+1$
Since $A$ contains $n+1$ elements, it follows that there is only one subset of $A$ containing $n+1$ elements(subset is $A$ itself). Since $\binom (n+1) k = \binom (n+1) (n+1) = 1$, it follows number of subsets of $A$ of $k = n+1$ elements is same as $\binom (n+1) k$.

• Case 3: $% $
There are two types of $k$-sized subsets of $A$. One type of subsets contain $a$ and other type does not contain $a$.

• Subsets that does not contain $a$ are also the subsets of $A'$. Now since $A'$ has $n$ elements and $% $, it follows by induction hypothesis that number of such subsets are $\binom n k$.

• Subsets that contains $a$ are of the form of $X \cup \{a\}$ where $X$ is $k-1$-sized subsets of $A'$. Thus number of $k$-sized subsets of $A$ that contains $a$ is equal to number of $k - 1$ sized subsets of $A'$. Since $A'$ contains $n$ elements, it follows from induction hypothesis that the number of subsets is $\binom n {k-1}$.

Thus total number of $k$-sized subsets of $A$ are ${ \binom n k } + {\binom n {k-1} }$. From part(b) this sum is equivalent to $\binom {n+1} k$.

Thus from all the cases, total number of $k$ sized subsets of $A$ is $\binom {n+1} k$.

(d)

Base Case:

For $n = 0$, LHS = ${(x+y)}^0 = 1$. And RHS = $x^0 \cdot y^0 = 1$. Thus $LHS = RHS$.

Induction Step:

Suppose theorem is correct for $n$. Thus ${(x+y)}^n = \sum_{k=0}^n {\binom {n} k} x^{n-k}y^k$.

For $n+1$, we have ${(x+y)}^{n+1} = (x+y){(x+y)}^n$. Thus by induction hypothesis:
$= (x+y)\sum_{k=0}^n {\binom {n} k} x^{n-k}y^k$
$= (x+y)( { {\binom {n} {0} }x^{n-0}y^{0} } + { {\binom {n} {1} }x^{n-1}y^{1} } + { {\binom {n} {2} }x^{n-2}y^{2} } + { {\binom {n} {3} }x^{n-3}y^{3} } + \cdot \cdot \cdot + { {\binom {n} {n} }x^{n-n}y^{n} } )$
$= (x+y)( { {\binom {n} {0} }x^{n} } + { {\binom {n} {1} }x^{n-1}y^{1} } + { {\binom {n} {2} }x^{n-2}y^{2} } + { {\binom {n} {3} }x^{n-3}y^{3} } + \cdot \cdot \cdot + { {\binom {n} {n} }y^{n} } )$

Multiplying each term inside by $(x+y)$ and using $\binom n 0 = \binom n n = 1$,
$= { x^{n+1} } + { {\binom {n} {0} }x^{n}y{1} } + { {\binom {n} {1} }x^{n}y^{1} } + { {\binom {n} {1} }x^{n-1}y^{2} } + { {\binom {n} {2} }x^{n-1}y^{2} } +{ {\binom {n} {2} }x^{n-2}y^{3} } + { {\binom {n} {3} }x^{n-2}y^{3} } + { {\binom {n} {3} }x^{n-3}y^{4} } + \cdot \cdot \cdot + { {\binom {n} {n} }x^{1}y^{n} } + { y^{n+1} }$
$\, = { x^{n+1} } + ({\binom {n} {0} } + {\binom {n} {1} })x^{n}y^{1} + ({\binom {n} {1} } + {\binom {n} {2} })x^{n-1}y^{2} + ({\binom {n} {2} } + {\binom {n} {3} })x^{n-2}y^{3} + ({\binom {n} {3} } + {\binom {n} {4} })x^{n-3}y^{4} + \cdot \cdot \cdot + ({\binom {n} {n-1} } + {\binom {n} {n} })x^{1}y^{n} + { y^{n+1} } \,$
$= { \binom {n+1} {0} }x^{n+1} + {\binom {n+1} {1} }x^{n}y^{1} + {\binom {n+1} {2} }x^{n-1}y^{2} + {\binom {n+1} {3} }x^{n-2}y^{3} + {\binom {n+1} {4} }x^{n-3}y^{4} + \cdot \cdot \cdot + {\binom {n+1} {n} }x^{1}y^{n} + {\binom {n+1} {n+1} }y^{n+1}$
$= \sum_{k=0}^{n+1} {\binom {n+1} k} x^{n+1-k}y^k$
Thus if $P(n)$ is true, then $P(n+1)$ is also true.

Soln19

(a)

Putting $x = 1, y = 1$ in part(d) of Ex18, we get $(1+1)^n = \sum_{i=0}^n {\binom {n} k}1^{n-k}1^k$. Thus:
$\Rightarrow 2^n = \sum_{i=0}^n {\binom {n} k}$

(b)

Putting $x = 1, y = -1$ in part(d) of Ex18, we get $(1-1)^n = \sum_{i=0}^n {\binom {n} k}1^{n-k}{(-1)}^{k}$. Thus:
$\Rightarrow \sum_{i=0}^n {\binom {n} k} {(-1)}^{k} = 0$
$\Rightarrow \sum_{i=0}^n {(-1)}^{k} {\binom {n} k} = 0$

Soln20

It will be easier to prove $% $.

Clearly $a_n^2$ is positive, it follows $a_{n+1}$ is also positive.

Base Case:

For $n = 1$, $a_1 = a_0^2 + \frac 1 4 = \frac 1 4$. Clearly $% $.

Induction Step:

Suppose theorem is correct for $n$.

Now $a_{n+1} = a_n^2 + {\frac 1 4}$
$% $
$= {\frac 1 4} + {\frac 1 4}$
$= \frac 1 2$

Thus $% $. Or we can also say that $% $.

Soln21

Adding $1$ helped made LHS closer to RHS. TODO