The value of (1 × 2²+2 × 3²+…+100 ×(101)²)/(1² × 2+2² × 3+… .+100² × 101) is

Solution

Maths sir · 4 steps

Write the r-th term of top and bottom, sum each with standard formulas, and nearly everything cancels.

1. General terms. The r-th term on top is r(r+1)² = r³ + 2r² + r. The r-th term below is r²(r+1) = r³ + r². Both run from r = 1 to n = 100.
2. Sum the denominator. Use Σ r³ = n²(n+1)²/4 and Σ r² = n(n+1)(2n+1)/6. Taking n(n+1)/12 out leaves 3n² + 7n + 2 = (3n+1)(n+2).
3. Sum the numerator. Add Σ r = n(n+1)/2 as well. Taking n(n+1)/12 out leaves 3n(n+1) + 4(2n+1) + 6 = 3n² + 11n + 10 = (3n+5)(n+2).

   (Σ r(r+1)²)/(Σ r²(r+1)) = (n(n+1)(n+2)(3n+5)/12)/(n(n+1)(n+2)(3n+1)/12) = (3n+5)/(3n+1)

4. So. With n = 100 the value is 305/301, which is option (a).

The idea to keep. For a ratio of two special sums, find the general term of each, sum with the standard formulas and factor fully. The common factors cancel, so you never need the huge numbers.

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