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  • Let the sixth term in the binomial expansion of <img alt="\left(\sqrt{\log _2\left(10-3^3\right)}+\sqrt[5]{2^{\log _2 3}}\right)^{\mathrm{m}}" src="https://entrancecorner.oncodecogs.com/gif.la

Let the sixth term in the binomial expansion of \left(\sqrt{\log _2\left(10-3^3\right)}+\sqrt[5]{2^{\log _2 3}}\right)^{\mathrm{m}} , in the increasing powers of 2^{(\mathrm{x}-2) \log _2 3}

be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of A.P., then the sum of the squares of all possible values of x is

 

Option: 1

4


Option: 2

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Option: 3

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Option: 4

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Answers (1)

best_answer

\Rightarrow \text { Sixth Term, } T_{5+1}={ }^m C_5\left(10-3^x\right)^{\frac{m-5}{2}} 3^{x-2}=21

\text { So, } 2^m C_2={ }^m C_1+{ }^m C_3

$$ 2 \frac{m(m-1)}{2}=m+\frac{m(m-1)(m-2)}{6} $$ \Rightarrow m=2,7$ (But $m=2$ is inadmissible) $\Rightarrow m=7

\text { Now, } T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21

\begin{aligned} & \Rightarrow \frac{10 \cdot 3^x-\left(3^x\right)^2}{3^2}=1 \\ & \left(3^x\right)^2-10 \cdot 3^x+9=0 \\ & 3^x=9,1 \\ & \Rightarrow x=0,2 \end{aligned}

Sum of squeals of values of x=0^2+2^2=4

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Anam Khan

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