help me in solving

Question

evaluate : $$ \sum_{k=0}^\infty\binom{n}{2+4k} $$ I tried using pre formulated series of multinomial expansions but it doesnt help. Please give a solution to the problem, without using complex numbers

Answer 1

Here is another idea.

Let $E_r(n)=\sum_{k=0}^\infty\binom{n}{r+4k}$ for $r=0,1,2,3$ and note $E_0(0)=1, E_1(0)=E_2(0)=E_3(0)=0$

We have the recurrences based on the formation of Pascal's triangle: $$E_0(n+1)=E_3(n)+E_0(n), E_1(n+1)=E_0(n)+E_1(n), E_2(n+1)=E_1(n)+E_2(n), E_3(n+1)=E_2(n)+E_3(n)$$

It is easy to prove by induction that $$E_1(n)+E_2(n)+E_3(n)+E_4(n)=2^n$$ and for $n\ge 1,$$$ E_1(n)+E_3(n)=E_0(n)+E_2(n)=2^{n-1}$$

So for $n\ge 1,$ $$ E_3(n)=2^{n-1}-E_1(n) \text{ and } E_0(n)=2^{n-1}-E_2(n)$$

So we have $$E_1(n+1)=2^{n-1}-E_2(n)+E_1(n)$$

and

$$E_2(n+1)=E_2(n)+E_1(n)$$ which is $$E_1(n)=E_2(n+1)-E_2(n)$$

Whence $$E_2(n+2)-E_2(n+1)=2^{n-1}-E_2(n)+E_2(n+1)-E_2(n)$$

This becomes the linear recurrence

$$E_2(n+2)=2E_2(n+1)-2E_2(n)+2^{n-1}$$ and the auxiliary equation is $$x^2-2x+2=0$$ which has roots $x=1\pm i$

Solving the recurrence with a particular solution for the inhomogeneous version plus a general solution for the homogeneous part, and then checking boundary conditions, gives the same elements as other solutions, and it is no surprise that the this method gives components $(1\pm i)^n$ and a multiple of $2^n$.

Answer 2

Hint:

For integer $n>0,\binom nr=0$ for $r<0,r>n$

So, we need $$\sum_{r=0,4r+2<n}\binom n{4r+2}$$

Now, $$(1+y)^n=\sum_{r=0}^n\binom nry^r\ \ \ \ \ (1)\implies y^2(1+y)^n=\sum_{r=0}^n\binom nry^{r+2}\ \ \ \ \ (2)$$

If $y^4=1, y=\pm1,\pm i$

set $y=\pm1,\pm i$ in $(ii)$ and add

Answer 3

I've made this more an extended hint and tried to give a different way into understanding lab bhattacharjee's method. I think of this as twisting a power series to get the right cancellations.

If $\omega$ is a complex $q^{th}$ root of unity we have that the sum $$S=1+\omega^r+\omega^{2r}+\dots +\omega^{(q-1)r}$$is equal to $q$ or $0$ according to whether $r$ is a multiple of $q$ or not. We can use this to pick out every $q^{th}$ element of a power series.

Here we would do $$(1+1)^n+(1+\omega)^n+(1+\omega^2)^n+\dots+(1+\omega^{(q-1)r})^n$$

This would pick out terms $0,q,2q,3q \dots$

If you want to pick out the terms $k, q+k \dots$ you have to shift the power series to get the right cancellation. But the same idea works.

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Here you want to have $4^{th}$ roots of unity and shift so that the coefficient is non-zero only when $r\equiv 2 \bmod 4$

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