If the sum of the coefficients in the expansion of $(1+2x)^n$ is 6561, then what is the value of the greatest coefficient in the expansion?
Solution
Standard Result:If $n$ is even positive integer then the greatest coefficient in the Binomial Expansion of $(x+a)^n$ is ${n\choose \frac{n}{2}}$
Sum of the Coefficients in the expansion of $(1+2x)^n=(1+2\times1)^n=3^n$.But it is given that Sum of the Coefficients in the expansion of $(1+2x)^n$ is 6561.
So,$3^n=6561=2^{11}\implies n=8$
Using the Standard result,we get Greatest Coefficient in the expansion of $(1+2x)^8=2^8(x+\frac{1}{2})$ is $2^8 \times {8\choose \frac{8}{2}}$=$2^8 \times {8\choose 4}=2^8 \times \frac{8\times 7\times 6 \times 5}{4\times3\times2\times1}=17920$
But the Correct answer is $1792$.
Please point out my mistake.
Thank you
Best Answer
The coefficient of $x$ in $(1+2x)^8$ is $^8C_r 2^r$ -- not simply $^8C_r$. So it's maximum won't occur at usual maximum of binomial coefficients alone (which occurs at $r=n/2$).
One instead has to maximize the general term as $$ \dfrac{\text{coefficient of } T_{r+1}}{\text{coefficient of } T_r} > 1 $$
and solve for $r$.
I got this way that $^8C_6 2^6$ is the largest coefficient. Indeed $$ ^8C_6 2^6 = 1792$$