The sum of 112 - 122 - 132 - - - 202 is

Let 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; = n(n+1)(2n+1)6) = 10 × 7 × ...

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Byju's Answer

Standard XII

Mathematics

Sigma n2

The sum of 11...

Question

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

3485

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2555

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2885

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2485

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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

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