Let the sixth term in the binomial expansion | vedclass

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

Question

(B) The expansion is $\left((10-3^x)^{1/2} + (3^{x-2})^{1/5}\right)^m$. The general term is $T_{r+1} = {^mC_r} (10-3^x)^{(m-r)/2} (3^{x-2})^{r/5}$.Giv

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The expansion is $\left((10-3^x)^{1/2} + (3^{x-2})^{1/5}\right)^m$. The general term is $T_{r+1} = {^mC_r} (10-3^x)^{(m-r)/2} (3^{x-2})^{r/5}$.Given $T_6 = 21$,so $r=5$: ${^mC_5} (10-3^x)^{(m-5)/2} (3^{x-2}) = 21$. Given ${^mC_1}, {^mC_2}, {^mC_3}$ are in $A.P.$,so $2({^mC_2}) = {^mC_1} + {^mC_3}$.$2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6} \implies m^2-m = m + \frac{m^3-3m^2+2m}{6}$.$6m^2-6m = 6m + m^3-3m^2+2m \implies m^3-9m^2+14m = 0$. Since $m \ge 5$,$m^2-9m+14=0 \implies (m-7)(m-2)=0$,so $m=7$.Substituting $m=7$ into $T_6$: ${^7C_5} (10-3^x)^{(7-5)/2} (3^{x-2}) = 21 \implies 21 (10-3^x) \cdot \frac{3^x}{9} = 21$.$(10-3^x) \cdot 3^x = 9 \implies 10 \cdot 3^x - (3^x)^2 = 9 \implies (3^x)^2 - 10 \cdot 3^x + 9 = 0$.$(3^x-9)(3^x-1) = 0 \implies 3^x=9$ or $3^x=1 \implies x=2$ or $x=0$.The sum of the squares of the values is $0^2 + 2^2 = 4$.

Similar Questions

Find the coefficient of $a^{4}$ in the product $(1+2a)^{4}(2-a)^{5}$ using the binomial theorem.

$\sum_{\substack{i, j=0 \\ i \neq j}}^{n} {}^{n}C_{i} {}^{n}C_{j}$ is equal to

Let $R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R]$,where $[x]$ denotes the greatest integer less than or equal to $x$,then $R f=$

The remainder when the polynomial $1+x^2+x^4+x^6+\ldots+x^{22}$ is divided by $1+x+x^2+x^3+\ldots+x^{11}$ is

If the coefficient of $x^r$ in the expansion of $(1+x+x^2+x^3)^{100}$ is $a_r$,and $S = \sum_{r=0}^{300} a_r$,then $\sum_{r=0}^{300} r \cdot a_r =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module