Question

If $s_n=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....\frac{1}{n},(n\in N)$, then $s_1+s_2+s_3+s_4+......s_n=(n+\lambda )s_{n+1}-(n+1)$. Find the value of $\lambda$.

Answer 1

$s_1=1$
$s_2=1+\frac{1}{2}$
$s_3=1+\frac{1}{2}+\frac{1}{3}$
$s_4=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$
.........................
$s_n=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+......\frac{1}{n-1}+\frac{1}{n}$
on adding
$s_1+s_2+.....s_n$
$n+\frac{n-1}{2}+\frac{n-2}{3}+\frac{n-3}{4}+.....\frac{n-(n-2)}{n-1}+\frac{n-(n-1)}{n}=n(1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{n})-(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+....\frac{n-2}{n-1}+\frac{n-1}{n})$
$={ns}_n-\left(\underset{(n-1)times}{\underset{}{1+1+1+....1}}\right)+(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....\frac{1}{n-1}+\frac{1}{n})$
$={ns}_n-(n-1)+s_n-1$
$=(n+1)s_n-n$