sumlimits_{m = 1}^n {{{tan }^{ - 1}}} left( { | vedclass

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

Question

(A) આપેલ છે કે $\sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} ight)} $$= \sum\limits_{m = 1}^n {{{	an }^{ - 1}}\

Vedclass · Accepted Answer

${	an ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} ight)$

Vedclass · Answer

(A) આપેલ છે કે $\sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} ight)} $$= \sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{2m}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} ight)} $$= \sum\limits_{m = 1}^n {{{	an }^{ - 1}}\left( {\frac{{({m^2} + m + 1) - ({m^2} - m + 1)}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} ight)} $$= \sum\limits_{m = 1}^n {[{{	an }^{ - 1}}({m^2} + m + 1) - {{	an ^{ - 1}}({m^2} - m + 1)]} }$$= ({	an ^{ - 1}}3 - {	an ^{ - 1}}1) + ({	an ^{ - 1}}7 - {	an ^{ - 1}}3) + ({	an ^{ - 1}}13 - {	an ^{ - 1}}7) + \dots + [{	an ^{ - 1}}({n^2} + n + 1) - {	an ^{ - 1}}({n^2} - n + 1)]$$= {	an ^{ - 1}}({n^2} + n + 1) - {	an ^{ - 1}}1$$= {	an ^{ - 1}}\left( {\frac{{({n^2} + n + 1) - 1}}{{1 + ({n^2} + n + 1)(1)}}} ight) = {	an ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} ight)$.

$\sum\limits_{m = 1}^n {{{\tan }^{ - 1}}} \left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} \right)$ ની કિંમત શોધો.

Similar Questions

$\cot \left(\sum_{n=1}^{23} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right)$ નું મૂલ્ય શોધો.

જો $4 \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$ હોય,તો $x =$ . . . . . . .

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{a - x}{1 + ax} \right) \right] = $

જો $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$ હોય,તો $x$ =

$\sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3}$ નો ઉકેલ શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module