If the sum of the coefficients of all the positive powers of x, in the Binomial expansion …

Question

If the sum of the coefficients of all the positive powers of $x$, in the Binomial expansion of $\left(x^n+\dfrac{2}{x^5}\right)^7$ is 939, then the sum of all the possible integral values of $n$ is\_\_\_\_

Answer: 57

Accepted Answer

$T_{r+1}=\binom7r x^{n(7-r)}2^r x^{-5r}=\binom7r2^r\,x^{7n-r(n+5)}$; positive power requires $r<\dfrac{7n}{n+5}$.

Testing $r=0,1,2,3,4$: sums of coefficients accumulate as $1,15,99,379,939$. So $r$ ranges $0$ to $4$, requiring $4<\dfrac{7n}{n+5}<5$:

$$4n+20<7n<5n+25\quad\Rightarrow\quad n>\frac{20}{3}\approx6.67\quad\text{and}\quad n<12.5$$

Integer values: $n=7,8,9,10,11,12$. Sum $=7+8+9+10+11+12=57$

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