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

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(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

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$10$

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(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 $n=$

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