Skip to content
Question of 114

Q.a) Find the nthn^{\text{th}} term of the sequence 3,5,7,…3, 5, 7, \ldots

(1)
b) Find the sum to nn terms of the series 3×12+5×22+7×32+…3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + \ldots (3)
Kerala DhseKerala DHSE Plus One Board 2018Subjective· 4mImportance★★★★★
0% · 0/114 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The odd-number sequence has nth term 2n+1; the kth term of the series is (2k+1)k² = 2k³ + k², summed using Σk³ and Σk² formulas.

a) 3, 5, 7, … is an AP with first term 3 and common difference 2, so nth term = 3 + (n − 1)·2 = 2n + 1.

b) The series 3·1² + 5·2² + 7·3² + … has kth term (2k + 1)·k² = 2k³ + k².

Sum to n terms = 2·Σk³ + Σk² = 2·[n²(n+1)²/4] + [n(n+1)(2n+1)/6]

= n²(n+1)²/2 + n(n+1)(2n+1)/6.

Factor out n(n+1)/6: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.