If (1³ + 2³ + 3³ + ... up to n terms)/(1 . 3 + 2 . 5 + 3 . 7 + ... up to n terms) = 9/5, then the value of n is

Solution

Maths sir · 5 steps

Write the general term of each series, sum both with the standard formulas, and n(n+1) cancels nicely.

1. Numerator. The sum of the first n cubes is (n(n+1)/2)² = n²(n+1)²/4.
2. Denominator. The first factors run 1, 2, 3, … and the second factors run 3, 5, 7, …, so the rth term is r(2r+1) = 2r² + r. Then Σ (2r² + r) = 2n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(4n+5)/6.
3. Divide and cancel. One factor of n(n+1) cancels, leaving n(n+1)/4 × 6/(4n+5).

   3n(n+1)/2(4n+5) = 9/5

4. Solve the quadratic. Cross-multiplying gives 15n² + 15n = 72n + 90, so 5n² - 19n - 30 = 0. This factorises as (n - 5)(5n + 6) = 0.
5. So. n must be a positive integer, so n = 5. Check: (3 × 5 × 6)/(2 × 25) = 90/50 = 9/5.

The idea to keep. For a series of products, first find the rth term as a polynomial in r. Then use Σ r, Σ r², Σ r³ and take out common factors like n(n+1) early.

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