Sum of squares of first n natural numbers=n(n+1)(2n+1)/6

1.	Sum of squares of first n natural numbers=n(n+1)(2n+1)/6
Answer» To find the sum of squares, we use the formula of (n+1)³ as shown in the picture.In the proof, I have used one result directly:1 + 2 + 3 + ... + n = n(n+1)/2Proof:Given series is an A.P. withFirst term = 1Last term = nNumber of terms = nSum = n/2 [First term + Last Term]So, Sum = n/2 [1 + n]So, Sum = n(n+1)/2That is, 1 + 2 + 3 + ... + n = n(n+1)/2\xa0

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