Let the sixth term in the binomial expansion of ( sqrt2^log _2(10 - 3^x) + sqrt[5]2^(x-2)log _2 3 )^m in the increasing powers of 2^(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 an A.P., then the sum of the squares of all possible values of x is __________.

^m C_1, ^m C_2, ^m C_3 are first, third and fifth term of A P Therefore a= ^m C_1 \ a+2 d= ^m C_2 \ a+4 d= ^m C_3 \ Therefore 2 ^m C_2- ^m C_3=m \ => m=7 or m=2 \ Because m=2 is not possible \ Therefore m=7 T_6= ^ m C_5(10-3^ x)^frac m-52 * (3^ x-2)=21 Putting value of m = 7, we get T_5+1= ^7 C_5(10-3^x)^(7-5 / 2) 3^x-2=21 \ => (10.3^x-(3^x)^2 / 3^2)=1\ => (3^x)^2-10 * 3^x+9=0 \ => 3^x=9,1 \ => x=0,2 Sum of squares of values of x = 0^(2) + 2^(2) = 4

Let the sixth term in the binomial expansion of 2log 2 ... | null (Mathematics)

{ }^m C_1,{ }^m C_2,{ }^m C_3

are first, third and fifth term of

A P

\begin{aligned} \therefore \quad & a={ }^m C_1 \\\\ & a+2 d={ }^m C_2 \\\\ & a+4 d={ }^m C_3 \\\\ \therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\ \Rightarrow & m=7 \text { or } m=2 \\\\ \because & m=2 \text { is not possible } \\\\ \therefore & m=7 \end{aligned}

\mathrm{T}_6={ }^{\mathrm{m}} \mathrm{C}_5\left(10-3^{\mathrm{x}}\right)^{\frac{\mathrm{m}-5}{2}} \cdot\left(3^{\mathrm{x}-2}\right)=21

Putting value of m = 7, we get

\begin{aligned} & T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21 \\\\ & \Rightarrow \frac{10.3^x-\left(3^x\right)^2}{3^2}=1\end{aligned}

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

Sum of squares of values of x = 0² + 2² = 4

Explanation

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