Prove that the sum of a square function is ((n(n+1)(2n+1))/6)


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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
	$p(n): 1^2 +2^2 +3^2 +...+n^2 = ((n(n+1)(2n+1))/6) •p(1): 1^2 =((1(1+1)( 2(1)+1 ))/6) 1=((1×2×3)/6)=1 ∴p(1) is true. •Let p(k) is true 1^2 +2^2 +3^2 +...+k^2 =((k(k+1)(2k+1))/6) p(k+1): 1^2 +2^2 +3^2 +...+k^2 +(k+1)^2 =((k(k+1)(2k+1))/6)+(k+1)^2 =((k(k+1)(2k+1)+6(k+1)^2 )/6) =(((k+1){k(2k+1)+6(k+1)})/6) =(((k+1)(2k^2 +7k+6))/6) =(((k+1)(k+2)(2k+3))/6) =(((k+1^(−) )(k+1^(−) +1)( 2(k+1^(−) )+1))/6) •∵ { ((p(1) is true)),(( p(k) is true⇒p(k+1) is true)) :} ∴ p(n) is true for n∈N$
	$p (n) :$ $1^{2} + 2^{2} + 3^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ $∙ p (1) :$ $1^{2} = \frac{1 (1 + 1) (2 (1) + 1)}{6}$ $1 = \frac{1 \times 2 \times 3}{6} = 1$ $∴ p (1) i s t r u e .$ $∙ L e t p (k) i s t r u e$ $1^{2} + 2^{2} + 3^{2} + . . . + k^{2} = \frac{k (k + 1) (2 k + 1)}{6}$ $p (k + 1) :$ $1^{2} + 2^{2} + 3^{2} + . . . + k^{2} + {(k + 1)}^{2} = \frac{k (k + 1) (2 k + 1)}{6} + {(k + 1)}^{2}$ $= \frac{k (k + 1) (2 k + 1) + 6 {(k + 1)}^{2}}{6}$ $= \frac{(k + 1) {k (2 k + 1) + 6 (k + 1)}}{6}$ $= \frac{(k + 1) (2 k^{2} + 7 k + 6)}{6}$ $= \frac{(k + 1) (k + 2) (2 k + 3)}{6}$ $= \frac{(\overset{―}{k + 1}) (\overset{―}{k + 1} + 1) (2 (\overset{―}{k + 1}) + 1)}{6}$ $∙ ∵ {\begin{cases} p (1) i s t r u e \\ p (k) i s t r u e \Rightarrow p (k + 1) i s t r u e \end{cases}$ $∴ p (n) i s t r u e f o r n \in N$
	Commented by necx122 last updated on 04/May/24

	$T h a n k y o u s o m u c h . T h i s i s m a t h e m a t i c a l$ $i n d u c t i o n, h o w e v e r, i f t h e$ $q u e s t i o n w a s j u s t t o p r o v e t h e s u m o f$ $a s q u a r e f u n c t i o n w i t h o u t k n o w i n g$ $t h e v a l u e b e f o r e, h o w s h o u l d t h a t b e$ $d o n e .$
	Answered by mr W last updated on 05/May/24

	$w e k n o w :$ $\underset{k = 1}{\sum^{n}} k = 1 + 2 + 3 + . . . + n = \frac{n (n + 1)}{2}$ ${(k + 1)}^{3} - k^{3} = 3 k^{2} + 3 k + 1$ $\underset{k = 1}{\sum^{n}} [{(k + 1)}^{3} - k^{3}] = 3 \underset{k = 1}{\sum^{n}} k^{2} + 3 \underset{k = 1}{\sum^{n}} k + n$ ${(n + 1)}^{3} - 1 = 3 \underset{k = 1}{\sum^{n}} k^{2} + 3 \times \frac{n (n + 1)}{2} + n$ $3 \underset{k = 1}{\sum^{n}} k^{2} = n^{3} + 3 n^{2} + 2 n - \frac{3 n (n + 1)}{2} = \frac{n (n + 1) (2 n + 1)}{2}$ $\Rightarrow \underset{k = 1}{\sum^{n}} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$
	Commented by mr W last updated on 04/May/24

	$i n s i m i l a r y w a y y o u c a n g e t$ $\underset{k = 1}{\sum^{n}} k^{3}, \underset{k = 1}{\sum^{n}} k^{4}, \underset{k = 1}{\sum^{n}} k^{5}, . . .$
	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

	$S (n) = \underset{j = 1}{\sum^{n}} j^{2} = ?$ $Assuming S (n) = {a n}^{3} + {b n}^{2} + c n + d$ $n = 1, 2, 3, 4$ $(1) a + b + c + d = 1$ $(2) 8 a + 4 b + 2 c + d = 5$ $(3) 27 a + 9 b + 3 c + d = 14$ $(4) 64 a + 16 b + 4 c + d = 30$ $Solving this system leads to$ $a = \frac{1}{3} \land b = \frac{1}{2} \land c = \frac{1}{6} \land d = 0$ $\Rightarrow$ $S (n) = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} = \frac{2 n^{2} + 3 n^{2} + n}{6} = \frac{n (n + 1) (2 n + 1)}{6}$ $We can test this for the next values n$ $S (5) = 55$ $S (6) = 91$ $. . .$ $or with different values for the insertion$ $n = 3, 7, 12, 19$ $(1) 27 a + 9 b + 3 c + d = 14$ $(2) 343 a + 49 b + 7 c + d = 140$ $(3) 1728 a + 144 b + 12 + d = 650$ $(4) 6859 a + 361 b + 19 c + d = 2470$ $which leads to the exact same result for S (n)$
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