The sum of 112-122-132-202 is

The correct option is D 2485Let nbsp; S =(11)^2 + (12)^2 + (13)^2 + … + (20)^2 = ∑_n=1^20n^2 ∑_n=1^10n^2 = 20 × 21 × 416 10 × 11 × 216 (∵∑_n=1^nn^2 nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Sum of Infinite Terms of a GP

The sum of 11...

Question

The sum of $(11)^{2} + (12)^{2} + (13)^{2} + \dots + (20)^{2}$ is

3485

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

2555

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

2885

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

2485

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

Open in App

Solution

The correct option is D $2485$
Let
$S = (11)^{2} + (12)^{2} + (13)^{2} + \dots + (20)^{2}$
$= 20 \sum n = 1 n^{2} - 10 \sum n = 1 n^{2}$
$= \frac{20 \times 21 \times 41}{6} - \frac{10 \times 11 \times 21}{6} (∵ n \sum n = 1 n^{2} = \frac{n (n + 1) (2 n + 1)}{6})$
$= 10 \times 7 \times 41 - 5 \times 11 \times 7 = 7 (410 - 55) = 2485$

Suggest Corrections

Similar questions

(11^{2} + 12^{2} + 13^{2} + . . . . . 20^{2}) =

Q. The sum of

10^{2} + 11^{2} + 12^{2} + . . . . + 20^{2}

Q. Sum infinite terms of the series

{cot}^{- 1} (1^{2} + \frac{3}{4}) + {cot}^{- 1} (2^{2} + \frac{3}{4}) + {cot}^{- 1} (3^{2} + \frac{3}{4}) + \dots

Q. Sum to infinite terms of the series

c o t^{- 1} (1^{2} + \frac{3}{4}) + c o t^{- 1} (2^{2} + \frac{3}{4}) + c o t^{- 1} (3^{2} + \frac{3}{4}) + . . . . . . .

Q. The sum of the infinite terms of the series

{cot}^{- 1} (1^{2} + \frac{3}{4}) + {cot}^{- 1} (2^{2} + \frac{3}{4}) + {cot}^{- 1} (3^{2} + \frac{3}{4}) + \dots

is equal to

Join BYJU'S Learning Program

The sum of (11)2+(12)2+(13)2+…+(20)2 is

The correct option is D 2485Let S=(11)2+(12)2+(13)2+…+(20)2 =20∑n=1n2−10∑n=1n2 =20×21×416−10×11×216(∵n∑n=1n2=n(n+1)(2n+1)6) =10×7×41−5×11×7 =7(410−55)=2485

The sum of $(11)^{2} + (12)^{2} + (13)^{2} + \dots + (20)^{2}$ is