frac{1}{1 cdot 2} + frac{1}{2 cdot 3} + frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given series is $S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}$.Using the method of partial fractions,we can write $\frac{1}{k(k+1)} = \frac{1}{k} - \

Vedclass · Accepted Answer

$\frac{n}{n + 1}$

Vedclass · Answer

(B) The given series is $S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}$.Using the method of partial fractions,we can write $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.Substituting this into the sum,we get:$S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right)$.This is a telescoping series where all intermediate terms cancel out:$S_n = 1 - \frac{1}{n+1}$.Simplifying the expression,we get $S_n = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)}$ equals

Similar Questions

The $4^{th}$ term of a $H.P.$ is $\frac{3}{5}$ and $8^{th}$ term is $\frac{1}{3},$ then its $6^{th}$ term is

Let $a, b, c, d \in R^+$ such that $256 abcd \geq (a+b+c+d)^4$ and $3a + b + 2c + 5d = 11$. Then,the value of $a^3 + b + c^2 + 5d$ is equal to:

$8^{th}$ term of the series $2\sqrt{2} + \sqrt{2} + 0 + \dots$ will be (in $\sqrt{2}$)

Find the sum of an $AP$ of $40$ terms whose first and last terms are $80$ and $275$.

If $x \in (0, \frac{\pi}{4})$,then the expression $\frac{\cos x}{\sin^2 x(\cos x - \sin x)}$ cannot take the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module