Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + ....

Solution:

The given series is 3 + 7 + 13 + 21 + 31 + ..., can be written as

S = 3 + 7 + 13 + 21 + 31 + .... + aₙ ₋ ₁ + aₙ ....(1)

S = 3 + 7 + 13 + 21 + .... aₙ ₋ ₂ + aₙ ₋ ₁ + aₙ ....(2)

On subtracting the equation (2) from (1) , we obtain

S - S = [3 + 7 + 13 + 21 + 31 + .... + aₙ ₋ ₁ + aₙ] - [3 + 7 + 13 + 21 + .... aₙ ₋ ₂ + aₙ ₋ ₁ + aₙ]

0 = 3 + [(7 - 3) + (13 - 7) + (21 - 13) + .... + (aₙ - aₙ - 1)] - aₙ

0 = 3 + [4 + 6 + 8 + .... (n - 1) terms]- aₙ

aₙ = 3 + [4 + 6 + 8 + .... (n - 1) terms]

Now using the sum of A.P. formula,

aₙ = 3 + [(n - 1)/2] [2 x 4 + (n - 1 - 1) 2]

aₙ = 3 + [(n - 1)/2] [8 + 2n - 4]

aₙ = 3 + [(n - 1)/2] [2n + 4]

aₙ = 3 + (n - 1)(n + 2)

aₙ = 3 + (n² + n - 2)

a = n² + n + 1

Therefore,

∑ⁿₖ ₌ ₁ (a)_k = k² + k + 1

= ∑ⁿₖ ₌ ₁ k² + ∑ⁿₖ ₌ ₁ k + ∑ⁿₖ ₌ ₁ 1

Now, using summation formulas,

= [n (n + 1)(2n + 1)]/6 + [n (n + 1)]/2 + n

= n [(n +1)(2n + 1) + 3(n + 1) + 6]/6

= n [2n² + 3n + 1 + 3n + 3 + 6]/6

= n [2n² + 6n + 10]/6

= n/3 (n² + 3n + 5)

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 23

Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + ....

Summary:

The sum of the first n terms of the series 3 + 7 + 13 + 21 + 31 + .... is n/3 (n² + 3n + 5)