Find the sum to n terms52-62-72-202

The given series is 5^2+6^2+7^2+…… + 20^2 =(1^2+2^2+3^2+…… + 20^2) (1^2+2^2+3^2+4^2) =∑^20_n=1n^2 ∑^4_n=1n^2 =20(20+1)(40+1)/6 4(4+1)(8+1)/6 =20(21)(41)/6 2 ...

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Byju's Answer

Standard XII

Mathematics

Sequence

Find the sum ...

Question

Find the sum to n terms

$5^{2} + 6^{2} + 7^{2} + \dots + 20^{2}$

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Solution

The given series is

$5^{2} + 6^{2} + 7^{2} + \dots \dots + 20^{2}$

$= (1^{2} + 2^{2} + 3^{2} + \dots \dots + 20^{2}) - (1^{2} + 2^{2} + 3^{2} + 4^{2})$

$= \sum_{n = 1}^{20} n^{2} - \sum_{n = 1}^{4} n^{2}$

$= \frac{20 (20 + 1) (40 + 1)}{6} - \frac{4 (4 + 1) (8 + 1)}{6}$

$= \frac{20 (21) (41)}{6} - \frac{20 (9)}{6}$

$= \frac{20}{6} [21 \times 41 - 9] = \frac{861 - 9}{6} \times 20$

$= \frac{852}{6} \times 20 = 2840$

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Similar questions

Find the sum to n terms of each of the series in Exercises 1 to 7.

5^{2} + 6^{2} + 7^{2} + . . . + 20^{2}

Q. Solve

5^{2} + 6^{2} + 7^{2} + . . . + 20^{2}

Q. Using the given pattern, find the missing numbers.

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + 42^{2} = . . .^{2}

Q. Find the following sum

\frac{1}{2^{2} - 1} + \frac{1}{(4^{2} - 1)} + \frac{1}{6^{2} - 1} + . . . . . + \frac{1}{20^{2} - 1}

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Find the sum to n terms 52+62+72+…+202

The given series is 52+62+72+……+202 =(12+22+32+……+202)−(12+22+32+42) =∑20n=1n2−∑4n=1n2 =20(20+1)(40+1)6−4(4+1)(8+1)6 =20(21)(41)6−20(9)6 =206[21×41−9]=861−96×20 =8526×20=2840

Find the sum to n terms

$5^{2} + 6^{2} + 7^{2} + \dots + 20^{2}$