Binomial theorem for positive integral Index

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Question

If the sum of the coefficients in the expansion of $(1 + 2 x)^{n}$ is $6561$ , then the greatest coefficient in the expansion, is

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Solution

Sum of the coefficients in the given expansion=f(1)= $3^{n} = 6561 = 3^{8}$ then $n = 8 .$
We have to find the greatest coefficients. This means that we
have to determine the greatest term for $x = 1$ .
$∴ \frac{T_{r + 1}}{T_{r}} = \frac{9 - r}{r} \times 2 x$ [See Ex. 16]
$\Rightarrow \frac{T_{r + 1}}{T_{r}} = \frac{(9 - r) \times 2}{r}$ $[∵ x = 1]$
Now $\frac{T_{r + 1}}{T_{r}} > 1 \Rightarrow 18 - 2 r > r \Rightarrow r < 6$
Thus, $6 t h$ and $7 t h$ terms are greatest and are equal in
magnitude.
Now,
$\begin{array}{l} T_{6} =^{8} C_{5} (1)^{8 - 5} (2)^{5} [∵ x = 1] \\ \Rightarrow T_{6} =^{8} C_{5} \times 2^{5} = 1792 \end{array}$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

If the sum of the coefficients in the expansion of (1+2x)n is 6561, then the greatest coefficient in the expansion, is

Ready to Test Your Skills?

free study material

If the sum of the coefficients in the expansion of $(1 + 2 x)^{n}$ is $6561$ , then the greatest coefficient in the expansion, is