The greatest positive integer k, for which 49ᵏ + 1 is a factor of the sum
49¹²⁵ + 49¹²⁴ + ..... + 49² + 49 + 1, is:

Solution

Maths sir · 5 steps

Count the terms first: 126 of them split into two halves of 63, and that split is the factorisation.

1. Count the terms. The powers run from 49⁰ to 49¹²⁵, so the sum has 126 terms. That is an even number, so the terms split into a first half and a second half of 63 terms each.
2. Take out 49 power. The second half is 49⁶³ + 49⁶⁴ + … + 49¹²⁵. Taking out 49⁶³ leaves 1 + 49 + … + 49⁶², which is exactly the first half.

   S = (1 + 49 + … + 49⁶²) + 49⁶³(1 + 49 + … + 49⁶²) = (49⁶³ + 1)(1 + 49 + … + 49⁶²)

3. Check with GP formula. The GP sum gives S = (49¹²⁶ - 1)/48 = (49⁶³ + 1) × (49⁶³ - 1)/48. The second factor is an integer because 49⁶³ - 1 is divisible by 49 - 1 = 48.
4. Why not larger. A factor 49ᵏ + 1 splits off like this only when the 126 terms form an even number of equal blocks of k terms, that is, when 2k divides 126. The largest such k is 126 ÷ 2 = 63.
5. So. The greatest k is 63, option (c).

The idea to keep. For 1 + x + … + xⁿ⁻¹ with n even, split the terms into two equal halves: the sum is (x^(n/2) + 1) times the first half. Count terms from x⁰.

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