The coefficient of t^{50} in (1 + t^2)^{25}(1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.Expanding $(1 + t^2)^{25}

Vedclass · Accepted Answer

$1 + ^{25}C_5$

Vedclass · Answer

(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.Expanding $(1 + t^2)^{25} = \sum_{k=0}^{25} {^{25}C_k} (t^2)^k = \sum_{k=0}^{25} {^{25}C_k} t^{2k}$.To obtain $t^{50}$,we consider the product of terms from $(1 + t^2)^{25}$ and the second factor $(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$:$1$. Multiplying $1$ from the second factor with the term $t^{50}$ from $(1 + t^2)^{25}$ (where $k=25$): Coefficient = $^{25}C_{25} = 1$.$2$. Multiplying $t^{40}$ from the second factor with the term $t^{10}$ from $(1 + t^2)^{25}$ (where $k=5$): Coefficient = $^{25}C_5$.Other combinations like $t^{25} 	imes t^{25}$ are not possible because $(1 + t^2)^{25}$ only contains even powers of $t$.Thus,the total coefficient of $t^{50}$ is $1 + ^{25}C_5$.

The coefficient of $t^{50}$ in $(1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47})$ is -

Similar Questions

In the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$,the coefficient of $x^4$ is

If the $7^{th}$ term from the beginning in the binomial expansion of ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln x} \right)^9}$ for $x > 0$ is equal to $729$,then the possible value of $x$ is:

If the $(r + 1)^{th}$ term in the expansion of $\left( \sqrt[3]{\frac{a}{\sqrt{b}}} + \sqrt{\frac{b}{\sqrt[3]{a}}} \right)^{21}$ has the same power of $a$ and $b$,then the value of $r$ is

The coefficient of ${x^{-9}}$ in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is

Find the term independent of $x$ $(x > 0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module