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

Binomial Theorem — JEE Main · 2022 · 27 Jun 2022 · Shift 2 Question with Solution

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

(x^{n} + \frac{2}{x ^{5}})^{7}

is 939, then the sum of all the possible integral values of n is :

Answer: 57

The general term in the expansion of

(x^{n} + \frac{2}{x ^{5}})^{7}

is:

T_{r + 1} = (r 7) (x^{n})^{7 - r} (\frac{2}{x ^{5}})^{r} = (r 7) 2^{r} x^{n (7 - r) - 5 r}

The power of

x

n (7 - r) - 5 r = 7 n - n r - 5 r = 7 n - (n + 5) r

.

For positive powers:

7 n - (n + 5) r > 0

, i.e.,

r < \frac{7 n}{n + 5}

.

The sum of coefficients of positive powers is given as 939.

The cumulative sum of coefficients:

So we need

r \leq 4

to give positive powers and

r = 5

to give the first non-positive power.

For

r = 4

7 n - 4 (n + 5) > 0 \Rightarrow 3 n - 20 > 0 \Rightarrow n \geq 7

For

r = 5

7 n - 5 (n + 5) < 0 \Rightarrow 2 n - 25 < 0 \Rightarrow n < 12.5

Since

n

is an integer:

7 \leq n \leq 12

.

Sum =

7 + 8 + 9 + 10 + 11 + 12 = 57

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer