The sum of n terms of the series 2²+5²+8²+ . . . . . . is

Correct Answer is : ( n2)(6n2+3n−1)≤ft(\;n/2)(6 n²+3 n-1)(2n)(6n2+3n−1)

Correct Answer is : ( n2)(6n2+3n−1)≤ft(\;n/2)(6 n²+3 n-1)(2n)(6n2+3n−1)

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Sequence and Series

Question : 71 of 86

Marks: +1, -0

The sum of

n

2^{2}+5^{2}+8^{2}+

. . . . . .

Solution:

Concept:

Sum of squares of terms of an arithmetic progression.

Explanation:

The given series is

2^2 + 5^2 + 8^2 + \cdots

The terms

2,5,8,\dots

are in AP with first term

2

and common difference

3

So the

k

‑th term is

2 + (k-1)3 = 3k-1

Thus the series is

\sum_{k=1}^{n} (3k-1)^2 = \sum_{k=1}^{n} (9k^2 - 6k + 1)

Using standard formulas:

\sum k = \frac{n(n+1)}{2}

\sum k^2 = \frac{n(n+1)(2n+1)}{6}

\sum 1 = n

S_n = 9\cdot\frac{n(n+1)(2n+1)}{6} - 6\cdot\frac{n(n+1)}{2} + n = \frac{3}{2}n(n+1)(2n+1) - 3n(n+1) + n

Factor

\frac{n}{2}

S_n = \frac{n}{2}\big[3(n+1)(2n+1) - 6(n+1) + 2\big]

Compute inside:

3(2n^2+3n+1) = 6n^2+9n+3

; subtract

6n+6

gives

6n^2+3n-3

; add

2

gives

6n^2+3n-1

Hence

S_n = \frac{n}{2}(6n^2+3n-1)

Answer:

\frac{n}{2}(6n^2+3n-1)

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