sumlimits_{r = 1}^{100} {frac{{tan ,{2^{r - 1 | vedclass

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

Question

(C) We know that $	an \, 2^r - 	an \, 2^{r-1} = \frac{\sin(2^r - 2^{r-1})}{\cos \, 2^r \cos \, 2^{r-1}} = \frac{\sin(2^{r-1})}{\cos \, 2^r \cos \, 2

Vedclass · Accepted Answer

$	an \, 2^{100} - 	an \, 1$

Vedclass · Answer

(C) We know that $	an \, 2^r - 	an \, 2^{r-1} = \frac{\sin(2^r - 2^{r-1})}{\cos \, 2^r \cos \, 2^{r-1}} = \frac{\sin(2^{r-1})}{\cos \, 2^r \cos \, 2^{r-1}} = \frac{	an \, 2^{r-1}}{\cos \, 2^r}$.Thus,the general term is $T_r = 	an \, 2^r - 	an \, 2^{r-1}$.Summing from $r=1$ to $100$ gives a telescoping series:$\sum\limits_{r=1}^{100} T_r = (	an \, 2^1 - 	an \, 2^0) + (	an \, 2^2 - 	an \, 2^1) + \dots + (	an \, 2^{100} - 	an \, 2^{99})$.All intermediate terms cancel out,leaving $	an \, 2^{100} - 	an \, 2^0$.Since $2^0 = 1$,the sum is $	an \, 2^{100} - 	an \, 1$.

$\sum\limits_{r = 1}^{100} {\frac{{\tan \,{2^{r - 1}}}}{{\cos \,{2^r}}}} $ is equal to

Similar Questions

If the sum of the series $\frac{1}{1 \cdot(1+d)} + \frac{1}{(1+d)(1+2d)} + \dots + \frac{1}{(1+9d)(1+10d)}$ is equal to $5$,then $50d$ is equal to:

What is the sum of $n$ terms of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots$?

If ${x_1}, {x_2}, {x_3}, \dots, {x_n}$ are in $A.P.$ whose common difference is $\alpha$,then the value of $\sin \alpha (\sec {x_1} \sec {x_2} + \sec {x_2} \sec {x_3} + \dots + \sec {x_{n-1}} \sec {x_n}) = $

If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$,then the maximum value of $a$ is.

If $a_1, a_2, \dots, a_n$ are in $A.P.$ with common difference $d$,then the sum of the series $\sin d (\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module