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The sum of squares of first n natural numbers is given by 1/6 n(n+1)(2n+1) or 1/6 (2n³+3n²+n). Find the sum of squares of the first 10 natural numbers

Solution:

Given, the sum of squares of first n natural numbers is given by 1/6 n(n + 1)(2n + 1) or 1/6 (2n³ + 3n² + n).

We have to find the sum of squares of the first 10 natural numbers.

Put n = 10 in the expression 1/6 n(n + 1)(2n + 1),

= 1/6 (10)(10 + 1)(20 + 1)

= 1/6 (10)(11)(21)

= (21/6)(110)

= (7/2)(110)

= 7(55)

= 385

Put n = 10 in the expression 1/6 (2n³ + 3n² + n)

= 1/6 [2(10)³ + 3(10)² + 10]

= 1/6 [2(1000) + 3(100) + 10]

= 1/6 [2000 + 300 + 10]

= 1/6 (2310)

= 385

Therefore, the sum of squares of the first 10 natural numbers is 385.

✦ Try This: The sum of first n natural numbers is given by 1/2 n² + 1/2 n³. Find the sum of the natural numbers from 21 to 30

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 12

NCERT Exemplar Class 7 Maths Chapter 10 Problem 87

The sum of squares of first n natural numbers is given by 1/6 n(n+1)(2n+1) or ⅙ (2n³+3n²+n). Find the sum of squares of the first 10 natural numbers

Summary:

The sum of squares of first n natural numbers is given by 1/6 n(n+1)(2n+1) or 1/6 (2n³+3n²+n). The sum of squares of the first 10 natural numbers is 385.

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