The sum of the first n terms of the series- 12-2-22-32-2-42-52-2-62- is n -2 n -12- when n is even- When n is odd- the sum is n2-2n-1B- -2n-12C- nn-1-2- n2-2n-1

The correct option is A When n is odd, 1^2 + 2.2^2 + 3^2 + 2.4^2 +…. + 2(n 1)^2 + n^2 = (n 1)n^2/2 + n^2, since (n 1) is even. = n^2/2 (n + 1) ...

You visited us 58 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of n Terms

The sum of th...

Question

The sum of the first n terms of the series, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + \dots . i s \frac{n}{2} (n + 1)^{2}$ , when n is even. When n is odd, the sum is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
When n is odd, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + \dots . + 2 (n - 1)^{2} + n^{2}$
$= \frac{(n - 1) n^{2}}{2} + n^{2}$ , since (n - 1) is even. = $\frac{n^{2}}{2} (n + 1)$

Suggest Corrections

Similar questions

Q. The sum of n terms of series

1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + . . .

\frac{n {(n + 1)}^{2}}{2}

when n is even. When n is odd, the sum is

Q. The sum of

n

terms of the series

1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + . . . .

\frac{n (n + 1)^{2}}{2}

when

n

is even. When

n

is odd, the sum is

Q. The sum of first

n

terms of the series

1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + . . . .

\frac{n (n + 1)^{2}}{2}

when

n

is even. When

n

is odd the sum is

Q. The sum of the first

n

terms of the series

1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + . . . . .

\frac{n (n + 1)^{2}}{2}

when n is even. When

n

is odd the sum is

The sum of first n terms of the given series $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + . . . . . . . . . i s \frac{n (n + 1)^{2}}{2}$ , When n is even. When n is odd, the sum will be

Join BYJU'S Learning Program

The sum of the first n terms of the series, 12+2.22+32+2.42+52+2.62+….isn2(n+1)2, when n is even. When n is odd, the sum is

The correct option is A When n is odd, 12+2.22+32+2.42+….+2(n−1)2+n2 =(n−1)n22+n2, since (n - 1) is even. = n22(n+1)

The sum of the first n terms of the series, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + \dots . i s \frac{n}{2} (n + 1)^{2}$ , when n is even. When n is odd, the sum is

The correct option is A
When n is odd, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + \dots . + 2 (n - 1)^{2} + n^{2}$
$= \frac{(n - 1) n^{2}}{2} + n^{2}$ , since (n - 1) is even. = $\frac{n^{2}}{2} (n + 1)$