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  • If the sum of the coefficient in the expansion of  is 6561, the greate

If the sum of the coefficient in the expansion of \small (1+2 x)^n is 6561, the greatest term in the expansion for x = 1/2 is 

Option: 1

4^{\text {th }}


Option: 2

5^{\text {th }}


Option: 3

6^{\text {th }}


Option: 4

None of these


Answers (1)

best_answer

Sum of the coefficients in the expansion of (1+2 x)^n=6561

\begin{aligned} & \Rightarrow(1+2 \mathrm{x})^{\mathrm{n}}=6561 \text {, when } \mathrm{x}=1 \\ & \Rightarrow 3^{\mathrm{n}}=6561 \Rightarrow 3^{\mathrm{n}}=3^8 \Rightarrow \mathrm{n}=8 \end{aligned}

Now, \frac{T_{r+1}}{T_r}=\frac{{ }^8 C_r(2 x)^r}{{ }^8 C_{r-1}(2 x)^{r-1}}=\frac{9-r}{r} \times 2 x

\Rightarrow \quad \frac{T_{r+1}}{T_r}=\frac{9-r}{r}                      \left[\because\ x=\frac{1}{2}\right]

\therefore \quad \frac{T_{r+1}}{T_r}>1 \Rightarrow \frac{9-r}{r}>1 \Rightarrow 9-r>r \Rightarrow 2 r<9 \Rightarrow r<4 \frac{1}{2}

Hence, 5^{th} term is the greatest term. 

Posted by

shivangi.bhatnagar

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