The sum of the coefficients of the first thre | vedclass

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

Question

(A) The coefficients of the first three terms of $(x - \frac{3}{x^2})^m$ are $^mC_0, -3(^mC_1)$,and $9(^mC_2)$.Given the condition: $^mC_0 - 3(^mC_1)

Vedclass · Accepted Answer

$-5940$

Vedclass · Answer

(A) The coefficients of the first three terms of $(x - \frac{3}{x^2})^m$ are $^mC_0, -3(^mC_1)$,and $9(^mC_2)$.Given the condition: $^mC_0 - 3(^mC_1) + 9(^mC_2) = 559$.Substituting the values: $1 - 3m + \frac{9m(m-1)}{2} = 559$.Multiplying by $2$: $2 - 6m + 9m^2 - 9m = 1118$.$9m^2 - 15m - 1116 = 0$.Dividing by $3$: $3m^2 - 5m - 372 = 0$.Solving the quadratic equation: $(m - 12)(3m + 31) = 0$.Since $m$ is a natural number,$m = 12$.The general term is $T_{r+1} = ^{12}C_r x^{12-r} (-3)^r x^{-2r} = ^{12}C_r (-3)^r x^{12-3r}$.For the term containing $x^3$,set $12 - 3r = 3$,which gives $r = 3$.The required term is $^{12}C_3 (-3)^3 x^3 = 220 	imes (-27) x^3 = -5940x^3$.

The sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x^2})^m$,where $x \neq 0$ and $m$ is a natural number,is $559$. Find the term of the expansion containing $x^3$. (in $x^3$)

Similar Questions

The number of integral terms in the expansion of $(3^{1/2} + 5^{1/4})^{680}$ is equal to

The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1:7:42$. Find $n$.

If $A$ and $B$ are coefficients of $x^{n}$ in the expansions of $(1+x)^{2n}$ and $(1+x)^{2n-1}$ respectively, then $A / B$ is equal to

If the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$ is $\frac{1}{6}$,then $n=$

If the sum of the coefficients of all the positive powers of $x$ in the binomial expansion of $(x^{n} + \frac{2}{x^{5}})^{7}$ is $939$,then the sum of all the possible integral values of $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module