proof-that-1-2-2-2-3-2-n-2-n-2n-1-n-1-6-

Question Number 96527 by student work last updated on 02/Jun/20

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

Commented by 06122004 last updated on 02/Jun/20

Answered by Farruxjano last updated on 02/Jun/20

Commented by student work last updated on 02/Jun/20

thanks dear great

Answered by Sourav mridha last updated on 02/Jun/20

t h e r e a r e many u s e f u l t e c h n i q u e s

b u t I w a n t t o s h o w v e r y

u n u s u a l m e t h o n d \dots

u s i n g E u l e r - M a c l a u r i n formula

\underset{m = 1}{\sum^{n}} f (m) = \int_{1}^{n} f (x) d x + \frac{1}{2} f (1) + \frac{1}{2} f (n)

+ \frac{B_{2}}{2!} [f^{^{'}} (n) - f^{^{'}} (1)] + \dots \dots \dots . .

w h e r e B_{n} = B e r n o u l l i n u m b e r

a n d here B_{2} = \frac{1}{6} \dots f (m) = m^{2} \dots s o

f (1) = 1 a n d f (n) = n^{2}

f (x) = x^{2} s o f^{^{'}} (n) = 2 n, f^{^{'}} (1) = 2

so \underset{m = 1}{\sum^{n}} m^{2} = \frac{n^{3}}{3} - \frac{1}{3} + \frac{1}{2} + \frac{n^{2}}{2} + \frac{(n - 1)}{6}

= \frac{2 n^{3} + 3 n^{2} + n}{6}

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

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