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

Another simple proof could be as follows: note that each square can be written as a sum of odd numbers: ∑k=1n(2k−1)=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(n−k+1)(2k−1)=(2n+3)∑k=1nk−(n+1)∑k=1n1−2S Therefore =(2n+3)n(n+1)2−n(n+1)=(2n+1)n(n+1)2

My contribution: ∑k=1nk2=14∑k=1n(2k)2 =14∑k=1n((2k),(2))+((2k+1),(2)) =14∑k=12n+1((k),(2)) =14((2n+2),(3)) =14⋅(2n+2)(2n+1)(2n)1⋅2⋅3=(n+1)(2n+1)n1⋅2⋅3 =16n(n+1)(2n+1)

Another simple proof could be as follows: note that each square can be written as a sum of odd numbers: ∑k=1n(2k−1)=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(n−k+1)(2k−1)=(2n+3)∑k=1nk−(n+1)∑k=1n1−2S Therefore =(2n+3)n(n+1)2−n(n+1)=(2n+1)n(n+1)2

My contribution: ∑k=1nk2=14∑k=1n(2k)2 =14∑k=1n((2k),(2))+((2k+1),(2)) =14∑k=12n+1((k),(2)) =14((2n+2),(3)) =14⋅(2n+2)(2n+1)(2n)1⋅2⋅3=(n+1)(2n+1)n1⋅2⋅3 =16n(n+1)(2n+1)

learning about definite integrals: sum_(k=1)^n k^2

eliaskidszs Answered2022-01-16

I encountered while learning about definite integrals:

\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 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 $\sum n$ )
Another argument is use:
$24 n^{2} + 2 = {(2 n + 1)}^{3} - {(2 n - 1)}^{3}$
and get a telescoping sum.
$26 \sum_{1}^{n} k^{2} + 2 n = \sum_{1}^{n} {(2 k + 1)}^{3} - \sum_{1}^{n} {(2 k - 1)}^{3}$
$24 \sum_{1}^{n} k^{2} + 2 n = {(2 n + 1)}^{3} - 1$
$24 \sum_{1}^{n} k^{2} = 8 n^{3} + 12 n^{2} + 4 n$
$24 \sum_{1}^{n} k^{2} = 4 n (n + 1) (2 n + 1)$
$\sum_{1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

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:
$\sum_{k = 1}^{n} (2 k - 1) = n^{2}$
When writing each square as a sum of odd numbers we get that
$S = \sum_{k = 1}^{n} k^{2} = 1 + (1 + 3) + (1 + 3 + 5) + \dots$
$= \sum_{k = 1}^{n} (n - k + 1) (2 k - 1) = (2 n + 3) \sum_{k = 1}^{n} k - (n + 1) \sum_{k = 1}^{n} 1 - 2 S$
Therefore
$= \frac{(2 n + 3) n (n + 1)}{2} - n (n + 1) = \frac{(2 n + 1) n (n + 1)}{2}$

alenahelenash

Expert

2022-01-24Added 366 answers

My contribution:

\sum_{k = 1}^{n} k^{2} = \frac{1}{4} \sum_{k = 1}^{n} (2 k)^{2}

= \frac{1}{4} \sum_{k = 1}^{n} ((2 k), (2)) + ((2 k + 1), (2))

= \frac{1}{4} \sum_{k = 1}^{2 n + 1} ((k), (2))

= \frac{1}{4} ((2 n + 2), (3))

= \frac{1}{4} \cdot \frac{(2 n + 2) (2 n + 1) (2 n)}{1 \cdot 2 \cdot 3} = \frac{(n + 1) (2 n + 1) n}{1 \cdot 2 \cdot 3}

= \frac{1}{6} n (n + 1) (2 n + 1)

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