I encountered while learning about definite integrals: ∑k=1nk2=n(n+1)(2n+1)6

eliaskidszs

eliaskidszs

Answered

2022-01-16

I encountered while learning about definite integrals:
k=1nk2=n(n+1)(2n+1)6

Answer & Explanation

Annie Levasseur

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 n )
Another argument is use:
24n2+2=(2n+1)3(2n1)3
and get a telescoping sum.
261nk2+2n=1n(2k+1)31n(2k1)3
241nk2+2n=(2n+1)31
241nk2=8n3+12n2+4n
241nk2=4n(n+1)(2n+1)
1nk2=n(n+1)(2n+1)6

abonirali59

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:
k=1n(2k1)=n2
When writing each square as a sum of odd numbers we get that
S=k=1nk2=1+(1+3)+(1+3+5)+
=k=1n(nk+1)(2k1)=(2n+3)k=1nk(n+1)k=1n12S
Therefore
=(2n+3)n(n+1)2n(n+1)=(2n+1)n(n+1)2

alenahelenash

alenahelenash

Expert

2022-01-24Added 366 answers

My contribution: k=1nk2=14k=1n(2k)2 =14k=1n((2k),(2))+((2k+1),(2)) =14k=12n+1((k),(2)) =14((2n+2),(3)) =14(2n+2)(2n+1)(2n)123=(n+1)(2n+1)n123 =16n(n+1)(2n+1)

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