Prove-that-the-sum-of-a-square-function-is-n-n-1-2n-1-6-

Question Number 207042 by necx122 last updated on 04/May/24

P r o v e t h a t t h e s u m o f a s q u a r e f u n c t i o n

i s \frac{n (n + 1) (2 n + 1)}{6}

Answered by Rasheed.Sindhi last updated on 04/May/24

p (n) :

1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

∙ p (1) :

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

1 = \frac{1 \times 2 \times 3}{6} = 1

∴ p (1) i s t r u e .

∙ L e t p (k) i s t r u e

1^{2} + 2^{2} + 3^{2} + \dots + k^{2} = \frac{k (k + 1) (2 k + 1)}{6}

p (k + 1) :

1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + {(k + 1)}^{2} = \frac{k (k + 1) (2 k + 1)}{6} + {(k + 1)}^{2}

= \frac{k (k + 1) (2 k + 1) + 6 {(k + 1)}^{2}}{6}

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

= \frac{(k + 1) (2 k^{2} + 7 k + 6)}{6}

= \frac{(k + 1) (k + 2) (2 k + 3)}{6}

= \frac{(\overset{―}{k + 1}) (\overset{―}{k + 1} + 1) (2 (\overset{―}{k + 1}) + 1)}{6}

∙ ∵ {\begin{cases} p (1) i s t r u e \\ p (k) i s t r u e \Rightarrow p (k + 1) i s t r u e \end{cases}

∴ p (n) i s t r u e f o r n \in N

Commented by necx122 last updated on 04/May/24

T h a n k y o u s o m u c h . T h i s i s m a t h e m a t i c a l

i n d u c t i o n, h o w e v e r, i f t h e

q u e s t i o n w a s j u s t t o p r o v e t h e s u m o f

a s q u a r e f u n c t i o n w i t h o u t k n o w i n g

t h e v a l u e b e f o r e, h o w s h o u l d t h a t b e

d o n e .

Commented by Rasheed.Sindhi last updated on 05/May/24

D {o n}^{'} t k n o w t h e w a y t o o b t a i n

t h e a b o v e f o r m u l a .

Answered by mr W last updated on 05/May/24

w e k n o w :

\underset{k = 1}{\sum^{n}} k = 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}

{(k + 1)}^{3} - k^{3} = 3 k^{2} + 3 k + 1

\underset{k = 1}{\sum^{n}} [{(k + 1)}^{3} - k^{3}] = 3 \underset{k = 1}{\sum^{n}} k^{2} + 3 \underset{k = 1}{\sum^{n}} k + n

{(n + 1)}^{3} - 1 = 3 \underset{k = 1}{\sum^{n}} k^{2} + 3 \times \frac{n (n + 1)}{2} + n

3 \underset{k = 1}{\sum^{n}} k^{2} = n^{3} + 3 n^{2} + 2 n - \frac{3 n (n + 1)}{2} = \frac{n (n + 1) (2 n + 1)}{2}

\Rightarrow \underset{k = 1}{\sum^{n}} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}

Commented by mr W last updated on 04/May/24

i n s i m i l a r y w a y y o u c a n g e t

\underset{k = 1}{\sum^{n}} k^{3}, \underset{k = 1}{\sum^{n}} k^{4}, \underset{k = 1}{\sum^{n}} k^{5}, \dots

Commented by necx122 last updated on 04/May/24

W o w!! I n e v e r t h o u g h t o f t h i s .

T h a n k y o u s i r .

Commented by mr W last updated on 04/May/24

Answered by Frix last updated on 05/May/24

S (n) = \underset{j = 1}{\sum^{n}} j^{2} = ?

Assuming S (n) = {a n}^{3} + {b n}^{2} + c n + d

n = 1, 2, 3, 4

(1) a + b + c + d = 1

(2) 8 a + 4 b + 2 c + d = 5

(3) 27 a + 9 b + 3 c + d = 14

(4) 64 a + 16 b + 4 c + d = 30

Solving this system leads to

a = \frac{1}{3} \land b = \frac{1}{2} \land c = \frac{1}{6} \land d = 0

\Rightarrow

S (n) = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} = \frac{2 n^{2} + 3 n^{2} + n}{6} = \frac{n (n + 1) (2 n + 1)}{6}

We can test this for the next values n

S (5) = 55

S (6) = 91

\dots

or with different values for the insertion

n = 3, 7, 12, 19

(1) 27 a + 9 b + 3 c + d = 14

(2) 343 a + 49 b + 7 c + d = 140

(3) 1728 a + 144 b + 12 + d = 650

(4) 6859 a + 361 b + 19 c + d = 2470

which leads to the exact same result for S (n)

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