Question

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 ____________.

Answer 1

Correct Answer: 57

 To find the sum of integral values of \( n \) that satisfy the given condition in the binomial expansion of \( \left(x^n+\frac{2}{x^5}\right)^7 \), we begin by considering the expansion terms:

• General term: \( T_r = \binom{7}{r} (x^n)^{7-r} \left(\frac{2}{x^5}\right)^r = \binom{7}{r} x^{n(7-r) - 5r} \cdot 2^r \)

We identify terms where the power of \( x \) is positive: \( n(7-r) - 5r > 0 \). Rearranging gives:

• \( n(7-r) > 5r \)
• \( n > \frac{5r}{7-r} \)

Next, we calculate the sum of coefficients of these terms:

• The sum of coefficients of positive powers means: \(\sum \binom{7}{r} 2^r\) for \( r \) such that \( n > \frac{5r}{7-r} \).

The sum of coefficients is given as 939:

• The binomial expansion of \( (1+2)^7 = 2187 \), full sum including all terms.
• Zero or negative powers of \( x \) yield a coefficient part which is: \(\sum \) for \( n ≤ \frac{5r}{7-r} \).

Let’s evaluate \( \left(x^n + \frac{2}{x^5}\right)^7 \) at \( x=1 \):

• Full sum: \( (1+2)^7 = 2187 \)

Subtract coefficients for \( n ≤ \frac{5r}{7-r} \), assume \( n > 5 \).

Find \( n \) that satisfy this condition:

• For \( r = 0 \) to \( 5 \): coefficient sum at \( n = 6 \) is feasible \( n > \frac{25}{2} \approx 12.5\), \( n = 13, 14...\).
• \( T_7 - \sum T_{6,5,...} = 939 \): possible when \( n = 6 \) only, determine \( n = 6 \).

Thus, the corresponding coefficient sum is 939, \( n = 6 \) aligns correctly within range 57:

Conclusion

The sum of all possible integral values of \( n \) is \( 6 \).