The sum of coefficients in ${(1 + x - 3{x^2})^{2134}}$ is
$-1$
$1$
$0$
${2^{2134}}$
(b) For, the sum of coefficients, put $x = 1$, to obtain the sum as ${(1 + 1 – 3)^{2134}} = 1$.
The value of $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + …..$ is equal to
If $C_{x} \equiv^{25} C_{x}$ and $\mathrm{C}_{0}+5 \cdot \mathrm{C}_{1}+9 \cdot \mathrm{C}_{2}+\ldots .+(101) \cdot \mathrm{C}_{25}=2^{25} \cdot \mathrm{k}$ then $\mathrm{k}$ is equal to
If the sum of the coefficients of all even powers of $x$ in the product $\left(1+x+x^{2}+\ldots+x^{2 n}\right)\left(1-x+x^{2}-x^{3}+\ldots+x^{2 n}\right)$ is $61,$ then $\mathrm{n}$ is equal to
$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + …. + 15\frac{{{C_{15}}}}{{{C_{14}}}} = $
$\frac{1}{{1!(n – 1)\,!}} + \frac{1}{{3!(n – 3)!}} + \frac{1}{{5!(n – 5)!}} + …. = $
Confusing about what to choose? Our team will schedule a demo shortly.
