If , then the value of is
Show answer and solution
Answer: 5
Solution
The numerator is [n(n+1)/2]² = n²(n+1)²/4.
For the denominator, the general term is k(2k+1) = 2k² + k, so the sum is 2·n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)/6 · [2(2n+1) + 3] = n(n+1)(4n+5)/6.
Setting up the equation:
[n²(n+1)²/4] / [n(n+1)(4n+5)/6] = 9/5
Simplify: [n(n+1)/4] · [6/(4n+5)] = 9/5
→ 6n(n+1) / [4(4n+5)] = 9/5
→ 30n(n+1) = 36(4n+5)
→ 30n² + 30n = 144n + 180
→ 30n² - 114n - 180 = 0
→ 5n² - 19n - 30 = 0
→ (5n + 6)(n - 5) = 0
Since n must be a positive integer, n = 5.
For the denominator, the general term is k(2k+1) = 2k² + k, so the sum is 2·n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)/6 · [2(2n+1) + 3] = n(n+1)(4n+5)/6.
Setting up the equation:
[n²(n+1)²/4] / [n(n+1)(4n+5)/6] = 9/5
Simplify: [n(n+1)/4] · [6/(4n+5)] = 9/5
→ 6n(n+1) / [4(4n+5)] = 9/5
→ 30n(n+1) = 36(4n+5)
→ 30n² + 30n = 144n + 180
→ 30n² - 114n - 180 = 0
→ 5n² - 19n - 30 = 0
→ (5n + 6)(n - 5) = 0
Since n must be a positive integer, n = 5.
[JEE Main 2023 · 24 Jan · Shift 2]