Find the sum of n terms of the series whose nth term is n2 + 3n

1.	Find the sum of n terms of the series whose nth term is n2 + 3n
Answer» t_n = n² + 3ⁿ ⇒ t₁ = 1² + 3¹ t₂ = 2² + 3² t³ = 3² + 3³ t⁴ = 4² + 3⁴ \(\vdots\)\(\vdots\) \(\vdots\) t_n = n²+ 3ⁿ ∴ Adding column wise, we get t₁ + t₂ + t₃ + ....... + t_n = (1² + 2² + 3² + ..... + n²) + (3¹ + 3² + ..... + 3ⁿ) = \(\displaystyle\sum_{k=1}^{n} k^2\) + \(\frac{3.(3^n-1)}{(3-1)}\) = \(\frac{n(n+1)(2n+1)}{6}\) + \(\frac{3}{2}\) (3ⁿ - 1).

Answer»

t_n = n² + 3ⁿ

⇒ t₁ = 1² + 3¹

t₂ = 2² + 3²

t³ = 3² + 3³

t⁴ = 4² + 3⁴

\(\vdots\)\(\vdots\) \(\vdots\)

t_n = n²+ 3ⁿ

∴ Adding column wise, we get

t₁ + t₂ + t₃ + ....... + t_n = (1² + 2² + 3² + ..... + n²) + (3¹ + 3² + ..... + 3ⁿ)

= \(\displaystyle\sum_{k=1}^{n} k^2\) + \(\frac{3.(3^n-1)}{(3-1)}\) = \(\frac{n(n+1)(2n+1)}{6}\) + \(\frac{3}{2}\) (3ⁿ - 1).

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