Question

prove that ⁿC₀aⁿb⁰+ⁿC₁aⁿ⁻¹b¹+ⁿC₂aⁿ⁻²b²+.......+ⁿCₙa⁰bⁿ=(a+b)ⁿ

Answer 1

Here's the proof by principle of mathematical induction.

We have to prove,

$\displaystyle P(n):\sum_{r=0}^n\left (\ ^n\!C_r\ a^{n-r}b^r\right)=(a+b)^n.$

Considering P(1),

$\text{LHS}=\ ^1\!C_0\ a^{1-0}b^0+\ ^1\!C_1\ a^{1-1}b^1=a+b\\\\\\\text{RHS}=(a+b)^1=a+b$

So assuming P(k) true,

$\displaystyle P(k):\sum_{r=0}^k\left (\ ^k\!C_r\ a^{k-r}b^r\right)=(a+b)^k.$

Consider P(k + 1).

$\displaystyle P(k+1):\sum_{r=0}^{k+1}\left (\ ^{k+1}\!C_r\ a^{k+1-r}b^r\right)=(a+b)^{k+1}.$

Let's prove this!

$\displaystyle\text {LHS}\\\\\\\sum_{r=0}^{k+1}\left (\ ^{k+1}\!C_r\ a^{k+1-r}b^r\right)$

Since $^nC_r+\ ^nC_{r+1}=\ ^{n+1}C_{r+1},$

$\displaystyle\sum_{r=0}^{k+1}\left(\left (\ ^k\!C_{r-1}+\ ^k\!C_r\right)a^{k+1-r}b^r\right)\\\\\\b\sum_{r=0}^{k+1}\left(\ ^kC_{r-1}\ a^{k-r+1}b^{r-1}\right)+a\sum_{r=0}^{k+1}\left (\ ^kC_r\ a^{k-r}b^r\right)\\\\\\b\left(\ ^kC_{-1}\ a^{k+1}b^{-1}\right)+b\sum_{r=1}^{k+1}\left(\ ^kC_{r-1}\ a^{k-r+1}b^{r-1}\right)+a\sum_{r=0}^{k}\left (\ ^kC_r\ a^{k-r}b^r\right)+a\left (\ ^kC_{k+1}\ a^{-1}b^{k+1}\right)$

Here the terms $b\left(\ ^kC_{-1}\ a^{k+1}b^{-1}\right)$ and $a\left (\ ^kC_{k+1}\ a^{-1}b^{k+1}\right)$ are not valid, so they're considered as 0. So,

$\displaystyle b\sum_{r=1}^{k+1}\left(\ ^kC_{r-1}\ a^{k-r+1}b^{r-1}\right)+a\sum_{r=0}^{k}\left (\ ^kC_r\ a^{k-r}b^r\right)$

To get it simpler and easily understandable,

$\displaystyle b\sum_{r=0}^{k}\left(\ ^kC_r\ a^{k-r}b^r\right)+a\sum_{r=0}^{k}\left (\ ^kC_r\ a^{k-r}b^r\right)$

From P(k),

$\displaystyle b(a+b)^k+a(a+b)^k\\\\\\(a+b)\cdot(a+b)^k\\\\\\(a+b)^{k+1}\\\\\\\text {RHS}$

Hence Proved!