If
(20)¹⁹+2(21)(20)¹⁸+3(21)²(20)¹⁷+…+20(21)¹⁹=k(20)¹⁹,
then k is equal to ___________.

Solution

Maths sir · 4 steps

Divide through by 20¹⁹ and you are left with an AGP in 21/20; shift and subtract.

1. Divide by power. Divide both sides by 20¹⁹ and put x = 21/20. You get k = 1 + 2x + 3x² + … + 20x¹⁹, an arithmetico-geometric series.
2. Shift and subtract. Multiply by x to get xk = x + 2x² + … + 19x¹⁹ + 20x²⁰. Subtract this from k. Each pair of matching powers leaves a coefficient of 1, and the last term 20x²⁰ is left over with a minus sign.

   (1-x)k = (1 + x + … + x¹⁹) - 20x²⁰ = (x²⁰-1)/(x-1) - 20x²⁰

3. Use x minus 1. Here x - 1 = 1/20, so the geometric part is 20(x²⁰ - 1) = 20x²⁰ - 20. The 20x²⁰ terms cancel and the right side is just -20.
4. So. (1-x)k = -20 with 1 - x = -1/20, so k = -20/(-1/20) = 400.

The idea to keep. When the coefficients go 1, 2, 3, … and the powers go up together, you have an AGP. Multiply by the ratio, subtract, and the series turns into a plain GP.

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