Q. 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:
Solution:
coefficients and there cumulative sum are:
Coefficient
Commulative sum
$x ^{7 n } \rightarrow{ }^{7} C _{0}$
1
$x ^{6 n -5} \rightarrow 2 \cdot{ }^{7} C _{1}$
1+14
$x ^{5 n -10} \rightarrow 2^{2} \cdot{ }^{7} C _{2}$
1 + 14+ 84
$ x ^{4 n -15} \rightarrow 2^{3} \cdot{ }^{7} C _{3}$
1+ 14+ 84+280
$ x ^{3 n -20} \rightarrow 2^{4} \cdot{ }^{7} C _{4}$
1+ 4+ 84+280+560 =939
$ x ^{2n -25} \rightarrow 2^{5} \cdot{ }^{7} C _{5}$
$3 n -20 \geq 0 \cap 2 n -25<0 \cap n \in I$
$\therefore 7 \leq n \leq 12$
Sum $=7+8+9+10+11+12=57$
| Coefficient | Commulative sum |
|---|---|
| $x ^{7 n } \rightarrow{ }^{7} C _{0}$ | 1 |
| $x ^{6 n -5} \rightarrow 2 \cdot{ }^{7} C _{1}$ | 1+14 |
| $x ^{5 n -10} \rightarrow 2^{2} \cdot{ }^{7} C _{2}$ | 1 + 14+ 84 |
| $ x ^{4 n -15} \rightarrow 2^{3} \cdot{ }^{7} C _{3}$ | 1+ 14+ 84+280 |
| $ x ^{3 n -20} \rightarrow 2^{4} \cdot{ }^{7} C _{4}$ | 1+ 4+ 84+280+560 =939 |
| $ x ^{2n -25} \rightarrow 2^{5} \cdot{ }^{7} C _{5}$ |