Question

If $c_0,c_1,c_2,......c_n$ are the coefficients in the expansion $(1+x{)}^n,$ where $n$ is a positive integer, shew that
$c_1-\frac{c_2}{2}+\frac{c_3}{3}-......+\frac{(-1{)}^{n-1}{c}_n}{n}=1+\frac{1}{2}+\frac{1}{3}+....\frac{1}{n}$

Answer 1

Let $S_n=c_1-\frac{c_2}{2}+\frac{c_3}{3}-\dots \dots +\frac{(-1{)}^{n-1}{c}_n}{n}$

We know the coefficients of the expansion of $(1+x{)}^n$

$\therefore S_n=n-\frac{n(n-1)}{2\cdot 2!}+\frac{n(n-1)(n-2)}{3\cdot 3!}-\dots \dots \text{to n terms}$

Similarly, $S_{n+1}=(n+1)-\frac{(n+1)n}{2\cdot 2!}-\frac{(n+1)n(n-1)}{3\cdot 3!}+\dots \dots \text{to n+1 terms}$

$S_{n+1}-S_n=1-\frac{n}{2!}+\frac{n(n-1)}{3!}-\dots \dots \text{to n+1 terms}$

Also we know that , $(1+x{)}^{n+1}=1+(n+1)x+\frac{(n+1)n}{2!}{x}^2+\frac{(n+1)n(n-1)}{3!}{x}^3+\dots \dots \text{to n+1 terms}$

For $x=-1,0=1-(n+1)+\frac{(n+1)n}{2!}-\frac{(n+1)n(n-1)}{3!}+\dots \dots \text{to n+1 terms}$

$\therefore 1-\frac{n}{2!}+\frac{n(n-1)}{3!}-\dots \dots \text{to n+1 terms}=\frac{1}{n+1}$

From above we can say that $S_{n+1}-S_n=\frac{1}{n+1}$

$⟹S_2-S_1=\frac{1}{2}$, but since $S_1=1$, thus $S_2=1+\frac{1}{2}$

Similarly, $S_3=S_2+\frac{1}{3}=1+\frac{1}{2}+\frac{1}{3}$

Similarly, $S_n=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots \dots +\frac{1}{n}$

Hence Proved