Questions

If ${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots . + a_{2 n} x^{2 n}$ , then

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a0−a2+a4−a6+…=0 ,if n is odd

a1−a3+a5−a7+…=0,if n is even

a0−a2+a4−a6+….0, if n=4p,p∈I+

a1−a3+a5−a7+….=0, if n=4p+1,p∈I+

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detailed solution

Correct option is A

∵1+x+x2n=a0+a1x+a2x2+…+a2nx2nPutting x=i(i=−1)Then, we get1+i+i2n=a0−a2+a4−a6+…+ia1−a3+a5−a7+…⇒ in=a0−a2+a4−a6+…+ia1−a3+a5−a7+…if n is odd, then Re(in) = 0⇒ a0−a2+a4−a6+…=0if n is even, then Im (in) = 0⇒ a1−a3+a5−a7+…=0

Similar Questions

For n $\geq$ 2, let $C_{r} = (_{r}^{n})$ and $a_{n} = {\sum_{r = 0}^{n}}_{} \frac{1}{C_{r}}$

Column-I	Column-II
A. $\sum_{r = 0}^{n} \frac{r}{C_{r}}$	P. ${na}_{n}$
B. $\sum_{r = 0}^{n} \frac{n - r}{C_{r}}$	Q. $\frac{1}{n} a_{n - 1}$
C. $\sum_{r = 0}^{' n} \frac{1}{^{r} C_{r}}$	R. $\frac{1}{2} {na}_{n}$
D. $\sum_{r = 0}^{n - 1} \frac{1}{(n - r) C_{r}}$	S. $\frac{1}{2 n} a_{n - 1}$