If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + \dots + a_{2n}$ is equal to

The sum of the series $\binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \binom{20}{3} + \dots + \binom{20}{10}$ is:

If the coefficients of $a^{r-1}$,$a^{r}$,and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression,prove that $n^{2}-n(4r+1)+4r^{2}-2=0$.

Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then the value of $C_0 + 2C_1 + 3C_2 + .... + (n + 1)C_n$ will be

If $C_r = { }^n C_r$,then find the sum $C_0 + C_4 + C_8 + C_{12} + \ldots$