STD 11 - 7. binomial theoram — JEE कक्षा 11 साइन्स — Question

$ = \frac{1}{2}[{(1 + x)^{4n}} + {(1 - x)^{4n}}]$

$x = 1$ एवं $x = i$ रखने पर,

$^{4n}{C_0} + {\,^{4n}}{C_2} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}} = \frac{1}{2}[{2^{4n}}]$

एवं $^{4n}{C_0} - {\,^{4n}}{C_2} + {\,^{4n}}{C_4} - ... + {\,^{4n}}{C_{4n}}$=$\frac{1}{2}[{(1 + i)^{4n}} + {(1 - i)^{4n}}]$

इस प्रकार,  $2{[^{4n}}{C_0} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}}]$

$ = {2^{4n - 1}} + \frac{1}{2}{[{(1 + i)^{4n}} + (1 - i)]^{4n}}$

अब, ${(1 + i)^{4n}} + {(1 - i)^{4n}} = {\left[ {\sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)} \right]^{4n}}$

$ + {\left[ {\sqrt 2 \left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right)} \right]^{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}$

$\therefore $$2{[^{4n}}{C_0} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}}] = {2^{4n - 1}} + \frac{1}{2}{2^{2n + 1}}{( - 1)^n}$

$⇒^{4n}{C_0} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}} = {2^{4n - 2}} + {( - 1)^n}{2^{2n - 1}}$

ट्रिक : $n = 1, 2$ रखकर जांच कीजिये