MCQ
$^{4n}{C_0}{ + ^{4n}}{C_4}{ + ^{4n}}{C_8} + ....{ + ^{4n}}{C_{4n}}$ = . . .
- ✓${2^{4n - 2}} + {( - 1)^n}{2^{2n - 1}}$
- B${2^{4n - 2}} + {2^{2n - 1}}$
- C${2^{2n - 1}} + {( - 1)^n}\,{2^{4n - 2}}$
- Dએકપણ નહિ.
$ = \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}}]$$ = {2^{4n - 1}} + \frac{1}{2}{[{(1 + i)^{4n}} + (1 - i)]^{4n}}$
Now, ${(1 + i)^{4n}} + {(1 - i)^{4n}} = {\left[ {\sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)} \right]^{4n}}$$ + {\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}$
==> $^{4n}{C_0} + {\,^{4n}}{C_4} + ... + {\,^{4n}}{C_{4n}} = {2^{4n - 2}} + {( - 1)^n}{2^{2n - 1}}$
Trick : Check by putting $n = 1, 2$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.