^{4n}{C_0}{ + ^{4n}}{C_4}{ + ^{4n}}{C_8} + .. | vedclass

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Question

(a) We have $^{4n}{C_0} + {\,^{4n}}{C_2}{x^2} + {\,^{4n}}{C_4}{x^4} + ... + {\,^{4n}}{C_{4n}}{x^{4n}}$

$ = \frac{1}{2}[{(1 + x)^{4n}} + {(1 - x)^{4

Vedclass · Accepted Answer

${2^{4n - 2}} + {( - 1)^n}{2^{2n - 1}}$

Vedclass · Answer

(a) We have $^{4n}{C_0} + {\,^{4n}}{C_2}{x^2} + {\,^{4n}}{C_4}{x^4} + ... + {\,^{4n}}{C_{4n}}{x^{4n}}$

$ = \frac{1}{2}[{(1 + x)^{4n}} + {(1 - x)^{4n}}]$

Putting $x = 1$ and $x = i$, we get

$^{4n}{C_0} + {\,^{4n}}{C_2} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}} = \frac{1}{2}[{2^{4n}}]$

and $^{4n}{C_0} - {\,^{4n}}{C_2} + {\,^{4n}}{C_4} - ... + {\,^{4n}}{C_{4n}}$=$\frac{1}{2}[{(1 + i)^{4n}} + {(1 - i)^{4n}}]$

Thus, $2{[^{4n}}{C_0} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}}]$$ = {2^{4n - 1}} + \frac{1}{2}{[{(1 + i)^{4n}} + (1 - i)]^{4n}}$

Now, ${(1 + i)^{4n}} + {(1 - i)^{4n}} = {\left[ {\sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} ight)} ight]^{4n}}$$ + {\left[ {\sqrt 2 \left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} ight)} ight]^{4n}}$

$ = {2^{2n}}(\cos n\pi + i\sin n\pi ) + {2^{2n}}(\cos n\pi - i\sin n\pi )$

$ = {2^{2n + 1}}\cos n\pi = {2^{2n + 1}}{( - 1)^n}$

$	herefore $$2{[^{4n}}{C_0} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}}] = {2^{4n - 1}} + \frac{1}{2}{2^{2n + 1}}{( - 1)^n}$

==> $^{4n}{C_0} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}} = {2^{4n - 2}} + {( - 1)^n}{2^{2n - 1}}$

Trick : Check by putting $n = 1, 2$.

$^{4n}{C_0}{ + ^{4n}}{C_4}{ + ^{4n}}{C_8} + ....{ + ^{4n}}{C_{4n}}$ = . . .

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