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 $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.Since $(1 + t^2)^{25} = \s

Vedclass · Accepted Answer

$1 + {}^{25}C_5$

Vedclass · Answer

(A) We need the coefficient of $t^{50}$ in the expansion of $(1 + t^2)^{25}(1 + t^{25} + t^{40} + t^{45} + t^{47} + \dots)$.Since $(1 + t^2)^{25} = \sum_{k=0}^{25} {}^{25}C_k (t^2)^k = \sum_{k=0}^{25} {}^{25}C_k t^{2k}$,the expansion only contains even powers of $t$.Thus,we only consider terms from the second part that are even powers of $t$ and less than or equal to $50$.The relevant terms are $1$ and $t^{40}$.For $1$,we need the coefficient of $t^{50}$ in $(1 + t^2)^{25}$,which is ${}^{25}C_{25} = 1$.For $t^{40}$,we need the coefficient of $t^{10}$ in $(1 + t^2)^{25}$,which is ${}^{25}C_5$.Therefore,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

The term independent of $x$ in the expansion of ${\left( {2x + \frac{1}{{3x}}} \right)^6}$ is

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $a/b$ equals:

If the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$ is $128$,then the greatest coefficient in the expansion of $(1 + x)^n$ is

The term independent of $x$ in the expansion of $(1-3x+2x^3)(\frac{3x^2}{2}-\frac{1}{3x})^9$ is

If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\frac{1}{x^3} - x^4\right)^n, x \neq 0$, is zero, then the value of $n$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module