If U_{n} = frac{1}{n} - frac{1}{n+1},then the | vedclass

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

Question

(B) Given $U_{n} = \frac{1}{n} - \frac{1}{n+1}$.This is a telescoping series.The sum $S_{5} = U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ can be written as

Vedclass · Accepted Answer

$\frac{5}{6}$

Vedclass · Answer

(B) Given $U_{n} = \frac{1}{n} - \frac{1}{n+1}$.This is a telescoping series.The sum $S_{5} = U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ can be written as:$S_{5} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{6})$Observing the terms,the intermediate fractions cancel out:$S_{5} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6}$$S_{5} = 1 - \frac{1}{6}$$S_{5} = \frac{6-1}{6} = \frac{5}{6}$

If $U_{n} = \frac{1}{n} - \frac{1}{n+1}$,then the value of $U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ is

Similar Questions

If $x=3$ and $y=4,$ then find the value of $256 x^{4}+160 x^{2} y^{2}+25 y^{4}$.

Find the values of $m$ and $n$ in the polynomial $2x^{3} + mx^{2} + nx - 14$,such that $(x-1)$ and $(x+2)$ are its factors.

If $a+b+c=0,$ then the value of $\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right)$ is:

Resolve into factors: $9(3 x+5 y)^{2}-12(3 x+5 y)(2 x+3 y)+4(2 x+3 y)^{2}$

If $x+y+z=0,$ then what is the value of $\frac{x^{2}}{3yz} + \frac{y^{2}}{3xz} + \frac{z^{2}}{3xy}?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module