For k ∈ ℕ, if the sum of the series 1+4/k+8/k²+13/k³+19/k⁴+… is 10 , then the value of k is _________.

Solution

Maths sir · 5 steps

The differences 3, 4, 5, 6 form an AP, so multiply by (1-x) twice and the series becomes geometric.

1. Spot the pattern. Put x = 1/k, so S = 1 + 4x + 8x² + 13x³ + 19x⁴ + … The numerators 1, 4, 8, 13, 19 have differences 3, 4, 5, 6, which go up by 1 each time. One subtraction leaves an AGP, so you subtract twice.
2. Subtract twice. Multiplying by (1-x) once gives S(1-x) = 1 + 3x + 4x² + 5x³ + … Doing it again gives S(1-x)² = 1 + 2x + x² + x³ + … = 1 + 2x + x²/(1-x). This is valid because k > 1 gives 0 < x < 1.

   S = ((1-x)(1+2x) + x²)/(1-x)³ = (1 + x - x²)/(1-x)³

3. Back to k. With x = 1/k, S = k(k² + k - 1)/(k-1)³. Setting this equal to 10 gives k³ + k² - k = 10(k-1)³, which simplifies to 9k³ - 31k² + 31k - 10 = 0.
4. Factorise. k = 2 satisfies it: 72 - 124 + 62 - 10 = 0. So the cubic is (k-2)(9k² - 13k + 5) = 0. The quadratic has discriminant 169 - 180 = -11 < 0, so it has no real roots.
5. So. The only natural number that works is k = 2. Check: with x = 1/2, S = 1.25/0.125 = 10. The answer is 2.

The idea to keep. Find the differences of the coefficients. If the first differences form an AP, multiply by (1-x) twice. The series then turns geometric and you can sum it.

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