Question

1+2+3....+n2 =n(n+1)(2n+1)/6
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Answer 1

1²+2²+3²+4²+...+n² = n(n+1)(2n+1) / 6

Consider the binomial expansion of:

(x–1)³ – x³ = –3x²+3x – 1

Putting values of x from 1 to n to obtain the partial sum

(1–1)³–(1)³ = –3(1)²+3(1)–1

(2–1)³–(2)³= –3(2)²+3(2)–1

... ... ... ... ...

(n–1)³–(n)³ = –3(n²)+3(n)–1

(+) (+) (+) (+) (+)

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–(n)³ = –(1²+2²+3²+...+n²) +3(1+2+3+...+n) –n

So, (1²+2²+3²+..+n²) = (n³+3(n²+n)/2 –n)/3

OR (1²+2²+3²+..+n²) = n(n+1)(2n+1) / 6