7.Binomial Theorem
hard

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

A

$47$

B

$57$

C

$67$

D

$87$

(JEE MAIN-2022)

Solution

coefficients and there cumulative sum are :

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

 

$3 n-20 \geq 0 \cap 2 n-25<0 \cap n \in I ~\\ \therefore \quad 7 \leq n \leq 12 ~\\ \text { Sum }=7+8+9+10+11+12=57$

Standard 11
Mathematics

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