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

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(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

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$	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

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