eliaskidszs

Answered

2022-01-16

I encountered while learning about definite integrals:

$\sum _{k=1}^{n}{k}^{2}=\frac{n(n+1)(2n+1)}{6}$

Answer & Explanation

Annie Levasseur

Expert

2022-01-16Added 30 answers

The standard method is induction and you should look it up as it is a popular second example (first is

Another argument is use:

and get a telescoping sum.

abonirali59

Expert

2022-01-17Added 35 answers

Another simple proof could be as follows: note that each square can be written as a sum of odd numbers:

When writing each square as a sum of odd numbers we get that

Therefore

alenahelenash

Expert

2022-01-24Added 366 answers

My contribution:
$\sum _{k=1}^{n}{k}^{2}=\frac{1}{4}\sum _{k=1}^{n}(2k{)}^{2}$
$=\frac{1}{4}\sum _{k=1}^{n}((2k),(2))+((2k+1),(2))$
$=\frac{1}{4}\sum _{k=1}^{2n+1}((k),(2))$
$=\frac{1}{4}((2n+2),(3))$
$=\frac{1}{4}\cdot \frac{(2n+2)(2n+1)(2n)}{1\cdot 2\cdot 3}=\frac{(n+1)(2n+1)n}{1\cdot 2\cdot 3}$
$=\frac{1}{6}n(n+1)(2n+1)$

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