Question

sa se calculeze suma
S=1+(1+2)+(1+2+3)+...+(1+2+3+..+n)
cu rezolvare ca la prosti daca puteti
​

Answer 1

$1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + n) = a _{n} \\ \\ a _{n} = \frac{1}{2} suma \: (n {}^{2} ) + suma \: (n) \\ \\ a _{n} = \frac{1}{2} \times \frac{n(n + 1)(2n + 1)}{6} + \frac{1}{2} \times \frac{n(n + 1)}{2} \\ \\ a _{n} = \frac{n(n + 1)(2n + 4)}{12}$

Answer 2

Explicație pas cu pas:

$\boxed{ {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6}}$

$\boxed {1+2+3+..+n = \frac{n(n + 1)}{2}}$

suma Gauss o putem scrie:

$1+2+3+..+n = \frac{n(n + 1)}{2} = \frac{ {n}^{2} + n}{2}=\\ = \frac{ {n}^{2}}{2} + \frac{n}{2}$

adică:

$1 = \frac{ {1}^{2} }{2} + \frac{1}{2}$

$1 + 2 = \frac{ {2}^{2} }{2} + \frac{2}{2}$

$1 + 2 + 3 = \frac{ {3}^{2} }{2} + \frac{3}{2}$

....

suma devine:

$S = 1+(1+2)+(1+2+3)+...+(1+2+3+..+n)$

$= \Big( \frac{ {1}^{2} }{2} + \frac{1}{2} \Big) + \Big( \frac{ {2}^{2} }{2} + \frac{2}{2} \Big) + \Big( \frac{ {3}^{2} }{2} + \frac{3}{2} \Big) + ... + \Big( \frac{ {n}^{2} }{2} + \frac{n}{2} \Big) =$

$= \frac{ {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2}}{2} + \frac{1 + 2 + 3 + ... + n}{2}$

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

$= \frac{1}{2} \cdot \frac{n(n + 1)(2n + 1) + 3n(n + 1)}{6}$

$= \frac{1}{2} \cdot \frac{n(n + 1)(2n + 4)}{6} = \frac{1}{2} \cdot \frac{2n(n + 1)(n + 2)}{6}$

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