If the sum of the coefficients of all the positive powers of x, in the binomial expansion of (xn+(2/x5))7 is 939 , then the sum of all the possible integral values of n is:

Question

If the sum of the coefficients of all the positive powers of x, in the binomial expansion of (xn+(2/x5))7 is 939 , then the sum of all the possible integral values of n is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

coefficients and there cumulative sum are:

Coefficient	Commulative sum
$x^{7 n} \to^{7} C_{0}$	1
$x^{6 n - 5} \to 2 \cdot^{7} C_{1}$	1+14
$x^{5 n - 10} \to 2^{2} \cdot^{7} C_{2}$	1 + 14+ 84
$x^{4 n - 15} \to 2^{3} \cdot^{7} C_{3}$	1+ 14+ 84+280
$x^{3 n - 20} \to 2^{4} \cdot^{7} C_{4}$	1+ 4+ 84+280+560 =939
$x^{2 n - 25} \to 2^{5} \cdot^{7} C_{5}$

3 n - 20 \geq 0 \cap 2 n - 25 < 0 \cap n \in I

∴ 7 \leq n \leq 12

Sum

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

Q. If the sum of the coefficients of all the positive powers of x, in the binomial expansion of (xn+x52​)7 is 939 , then the sum of all the possible integral values of n is:

Q. 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: