Displaystylefrac - 1 - 1times 2 - -frac - 1 - 2times 3 - -frac - 1 - 3times 4 - -

Question

Toppr Admin · Accepted Answer

The given series is-11-xD7-2-12-xD7-3-13-xD7-4-nth  term-an-1n-n-1-1n-x2212-1n-1-by partial fractions-x2234-a1-11-x2212-12a2-12-x2212-13a3-13-x2212-14-an-1n-x2212-1n-1Adding  the above terms columnwise- we obtaina1-a2-an-11-12-13-1n-x2212-12-13-14-1n-1-x2234-Sn-1-x2212-1n-1-n-1-x2212-1n-1-nn-1

$\frac{2}{3}$	$\frac{4}{3}$
$\frac{1}{3}$	$\frac{2}{3}$

$\frac{1}{2}$	$\frac{1}{3}$
$\frac{1}{3}$	$\frac{1}{4}$

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + . . . .$

$The given series is,11×2+12×3+13×4+...nth term,an=1n(n+1)=1n−1n+1,(by partial fractions)∴a1=11−12a2=12−13a3=13−14...an=1n−1n+1Adding the above terms columnwise, we obtaina1+a2+...+an=[11+12+13+...1n]−[12+13+14+...1n+1]∴Sn=1−1n+1=n+1−1n+1=nn+1$

11×2+12×3+13×4+....

The given series is,11×2+12×3+13×4+...nth term,an=1n(n+1)=1n−1n+1,(by partial fractions)∴a1=11−12a2=12−13a3=13−14...an=1n−1n+1Adding the above terms columnwise, we obtaina1+a2+...+an=[11+12+13+...1n]−[12+13+14+...1n+1]∴Sn=1−1n+1=n+1−1n+1=nn+1

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + . . . .$

$The given series is,11×2+12×3+13×4+...nth term,an=1n(n+1)=1n−1n+1,(by partial fractions)∴a1=11−12a2=12−13a3=13−14...an=1n−1n+1Adding the above terms columnwise, we obtaina1+a2+...+an=[11+12+13+...1n]−[12+13+14+...1n+1]∴Sn=1−1n+1=n+1−1n+1=nn+1$