Question

$\rm \red{ Prove \: that : - }\\ \displaystyle\rm \red{ {{{(a + b {)}^n = \sum_{k = 0}^{n} \binom{n}{k} {a}^{n - k} {b}^{k} }}}}$​

Answer 1

Appropriate Question :-

Prove that,

$\displaystyle\rm{ {{{(a + b {)}^n = \sum_{k = 0}^{n} \binom{n}{k} {a}^{n - k} {b}^{k}}}}} \: \: \forall \: n \: \in \: N$

$\large\underline{\sf{Solution-}}$

$\displaystyle\rm{ {{{(a + b {)}^n = \sum_{k = 0}^{n} \binom{n}{k} {a}^{n - k} {b}^{k}}}}}$

can be rewritten as

$\displaystyle\rm{ {{{(a + b {)}^n = \binom{n}{0} {a}^{n} {b}^{0}}}}} + \binom{n}{1}{a}^{n - 1} {b}^{1} + - - + \binom{n}{n}{a}^{0} {b}^{n}$

Now, we use Principal of Mathematical Induction, to prove this result.

Let assume that

$\rm \: P(n) : \displaystyle\rm{ {{{(a + b {)}^n = \binom{n}{0} {a}^{n} {b}^{0}}}}} + \binom{n}{1}{a}^{n - 1} {b}^{1} + - - + \binom{n}{n}{a}^{0} {b}^{n}$

Step :- 1 For n = 1

$\displaystyle\rm{ {{{(a + b {)}^1 = \binom{1}{0} {a}^{1} {b}^{0}}}}} + \binom{1}{1}{a}^{1 - 1} {b}^{1}$

$\rm \: a + b = a + b$

$\rm\implies \:P(n) \: is \: true \: for \: n = 1$

Step :- 2 Let assume that P(n) is true for n = r, where r is some natural number.

$\displaystyle\rm{ {{{(a + b {)}^r = \binom{r}{0} {a}^{r} {b}^{0}}}}} + \binom{r}{1}{a}^{r - 1} {b}^{1} + - - + \binom{r}{r}{a}^{0} {b}^{r}$

Step :- 3 We have to prove that P(n) is true for n = r + 1

$\displaystyle\rm{ {{{(a + b)^{r + 1} = \binom{r + 1}{0} {a}^{r + 1} {b}^{0}}}}} + \binom{r + 1}{1}{a}^{r} {b}^{1} + - - + \binom{r + 1}{r + 1}{a}^{0} {b}^{r + 1}$

Now, Consider

$\displaystyle\rm{ {{{(a + b)^{r + 1}}}}}$

can be rewritten as

$\rm \: = \:(a + b) {(a + b)}^{r}$

$\rm \: = \:(a + b)\bigg( \binom{r}{0} {a}^{r} {b}^{0}+ \binom{r}{1}{a}^{r - 1} {b}^{1} + - - + \binom{r}{r}{a}^{0} {b}^{r} \bigg)$

$\displaystyle\rm \: = \:\binom{r}{0}{a}^{r + 1} {b}^{0} + \binom{r}{1}{a}^{r} {b}^{1} + - - + \binom{r}{r}{a}^{1} {b}^{r} + \\ \rm \: \: \:\binom{r}{0}{a}^{r} {b}^{1} + \binom{r}{1}{a}^{r - 1} {b}^{2} + - - + \binom{r}{r}{a}^{0} {b}^{r + 1}$

$\displaystyle\rm \: = \:\binom{r}{0}{a}^{r + 1} {b}^{0} + \bigg[\binom{r}{1} + \binom{r}{0}\bigg]{a}^{r} {b}^{1} + - - + \bigg[\binom{r}{r} + \binom{r}{r - 1}\bigg]{a}^{1} {b}^{r} + + \binom{r}{r}{a}^{0} {b}^{r + 1}$

can also be rewritten as

$\displaystyle\rm \: = \:\binom{r + 1}{0}{a}^{r + 1} {b}^{0} + \binom{r + 1}{1}{a}^{r} {b}^{1} + - - + \binom{r + 1}{r}{a}^{1} {b}^{r} + + \binom{r + 1}{r + 1}{a}^{0} {b}^{r + 1}$

$\boxed{\sf{  \: \because \: \binom{r}{0} \: = \: \binom{r + 1}{0} \: \: }}$

$\boxed{\sf{  \: \because \: \binom{n}{r} + \binom{n}{r - 1}\: = \: \binom{n + 1}{r} \: \: }}$

$\boxed{\sf{  \: \because \: \binom{r}{r} \: = \: \binom{r + 1}{r + 1} \: \: }}$

Hence,

$\displaystyle\rm{ {{{(a + b)^{r + 1} = \binom{r + 1}{0} {a}^{r + 1} {b}^{0}}}}} + \binom{r + 1}{1}{a}^{r} {b}^{1} + - - + \binom{r + 1}{r + 1}{a}^{0} {b}^{r + 1}$

$\rm\implies \:P(n) \: is \: true \: for \: n = r + 1$

Hence, By the process of Principal of Mathematical Induction,

$\displaystyle\rm{ {{{(a + b {)}^n = \sum_{k = 0}^{n} \binom{n}{k} {a}^{n - k} {b}^{k}}}}} \: \: \forall \: n \: \in \: N$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{\sf{  \:\rm \: \binom{n}{0} \: = \: 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \binom{n}{1} \: = \: n \: \: }}$