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 _________.

Solution

Maths sir · 4 steps

Find which terms add up to 939 first, then choose n so exactly those terms carry positive powers.

1. General term. Tᵣ₊₁ = C(7, r) xⁿ⁽⁷⁻ʳ⁾ (2/x⁵)ʳ = C(7, r) 2ʳ x⁷ⁿ⁻ʳ⁽ⁿ⁺⁵⁾. The coefficient is C(7, r)2ʳ, and the power of x falls by n+5 each time r goes up by 1.

   r	Coefficient C(7, r)2ʳ	Running total
   0	1	1
   1	14	15
   2	84	99
   3	280	379
   4	560	939
   5	672	1611
   6	448	2059
   7	128	2187

2. Which terms. The total 939 is reached by the terms r = 0 to 4. So the powers must be positive for r ≤ 4 and not positive for r = 5. The powers only fall as r grows, so checking r = 4 and r = 5 covers every term.

   7n - 4(n+5) > 0 and 7n - 5(n+5) ≤ 0

3. Solve. The first condition gives 3n > 20, so n > 6.67. The second gives 2n ≤ 25, so n ≤ 12.5. The integer values are n = 7, 8, 9, 10, 11, 12.
4. So. The sum is 7 + 8 + 9 + 10 + 11 + 12 = 57.

The common slip: 63. Rounding 3n > 20 down to n ≥ 6 brings in n = 6 and gives 63. But at n = 6 the r = 4 power is 42 - 44 = -2, which is not positive.

The idea to keep. When the power of x changes by the same amount from term to term, the positive powers form one unbroken block of terms. Match the given coefficient sum to that block, then bound n using the two terms at its edge.

Answer from the NTA final answer key. Solution by PrepWiser.