If ( )/(4)+ _p=1^11 ^-1(2^p-11+2^2p-1)= , then is equal to ____ .

Question

If $\frac{\pi }{4}+\sum _{p=1}^{11}{	an }^{-1}(\frac{{2}^{p-1}}{1+{2}^{2p-1}})=\alpha $, then $	an \alpha $ is equal to $____$ .

Answer: 2048

Accepted Answer

Use the identity

${	an }^{-1}(\frac{x-y}{1+xy})={	an }^{-1}x-{	an }^{-1}y$

when $xy>-1$.

Now observe the given term:

$\frac{{2}^{p-1}}{1+{2}^{2p-1}}$

We want to write it in the form

$\frac{x-y}{1+xy}$

Take

$x={2}^{p},\, y={2}^{p-1}$

Then

$x-y={2}^{p}-{2}^{p-1}={2}^{p-1}$

and

$xy={2}^{p}\cdot {2}^{p-1}={2}^{2p-1}$

So,

$\frac{{2}^{p-1}}{1+{2}^{2p-1}}=\frac{{2}^{p}-{2}^{p-1}}{1+{2}^{p}\cdot {2}^{p-1}}$

Hence,

${	an }^{-1}(\frac{{2}^{p-1}}{1+{2}^{2p-1}})={	an }^{-1}({2}^{p})-{	an }^{-1}({2}^{p-1})$

Therefore,

$\sum _{11}^{p=1}{	an }^{-1}(\frac{{2}^{p-1}}{1+{2}^{2p-1}})=\sum _{11}^{p=1}[{	an }^{-1}({2}^{p})-{	an }^{-1}({2}^{p-1})]$

This is a telescoping sum:

$=({	an }^{-1}2-{	an }^{-1}1)+({	an }^{-1}4-{	an }^{-1}2)+\cdots +({	an }^{-1}{2}^{11}-{	an }^{-1}{2}^{10})$

So all middle terms cancel, and we get

$\sum _{11}^{p=1}{	an }^{-1}(\frac{{2}^{p-1}}{1+{2}^{2p-1}})={	an }^{-1}({2}^{11})-{	an }^{-1}(1)$

Since

${	an }^{-1}(1)=\frac{\pi }{4}$

given

$\alpha =\frac{\pi }{4}+\sum _{11}^{p=1}{	an }^{-1}(\frac{{2}^{p-1}}{1+{2}^{2p-1}})$

becomes

$\alpha =\frac{\pi }{4}+{	an }^{-1}({2}^{11})-\frac{\pi }{4}={	an }^{-1}({2}^{11})$

Thus,

$	an \alpha =	an ({	an }^{-1}({2}^{11}))={2}^{11}=2048$

So, the answer is

$2048$

Solutions · 1