If left(1+frac{3}{1}right)left(1+frac{5}{4}ri | vedclass

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

Question

(B) We have,$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2n+1}{n^2}\right)=121$ Simplifying eac

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) We have,$\left(1+\frac{3}{1}ight)\left(1+\frac{5}{4}ight)\left(1+\frac{7}{9}ight) \ldots\left(1+\frac{2n+1}{n^2}ight)=121$ Simplifying each term: $\left(\frac{4}{1}ight)\left(\frac{9}{4}ight)\left(\frac{16}{9}ight) \ldots\left(\frac{n^2+2n+1}{n^2}ight)=121$ This is a telescoping product: $\left(\frac{4}{1}ight) 	imes \left(\frac{9}{4}ight) 	imes \left(\frac{16}{9}ight) 	imes \ldots 	imes \left(\frac{(n+1)^2}{n^2}ight)=121$ Canceling the intermediate terms,we are left with: $\frac{(n+1)^2}{1} = 121$ $(n+1)^2 = 121$ $n+1 = 11$ $n = 10$

If $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2n+1}{n^2}\right)=121$,then $n=$

Similar Questions

$CrO_3$ dissolves in aqueous $NaOH$ to give:

If $\operatorname{adj}\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}$,then $\begin{bmatrix} a & b \end{bmatrix}$ is equal to

The equation of the line passing through the point of intersection of the lines $x-3y+2=0$ and $2x+5y-7=0$ and perpendicular to the line $3x+2y+5=0$ is:

Peptidoglycan is present in:

What is the genetic material of the Tobacco Mosaic Virus $(TMV)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module