User:Yunzhong Hou

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$n 3$

$\frac{1}{\tan c}$

$\sqrt{2 + 4/5 \frac{3}{2}}$

$\sum_{k=1}^N k^2 + k^3 + k^4 + ...$

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

$1(2) + 2(3) + 3(4) + ... + n (n + 1) = ?$

$a n = n (n + 1)$

$S n$

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

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

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

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

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

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

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

$12 + 2 2 + ... + n 2 = (1(2) + 2(3) + ... + n (n + 1)) + (1 + 2 + ... + n)$

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

$= \frac{1}{3} n^3 - \frac{1}{2} n^2 + \frac{1}{6} n$

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

$1(2)(3) = \frac{1(1+1)(1+2)(1+3)}{4} = 6$

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

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

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

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

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

$= \frac{1}{4}n^4- \frac{1}{2}n^3+ \frac{1}{4}n^2$

$= \frac{1}{4}n^4- \frac{1}{2}n^3+ \frac{1}{4}n^2$

$S_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}$

$1(2)(3)(4) = \frac{1(1+1)(1+2)(1+3)(1+4)}{5} = 24$

$1(2)(3)(4)+2(3)(4)(5)+...+n(n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}$

$1(2)(3)(4) + 2(3)(4)(5) + ... + n (n + 1)(n + 2)(n + 3) + (n + 1)(n + 2)(n + 3)(n + 4)$

$= \frac{n(n+1)(n+2)(n+3)(n+4)}{5}+(n+1)(n+2)(n+3)(n+4)$

$= \frac{n(n+1)(n+2)(n+3)(n+4)+5(n+1)(n+2)(n+3)(n+4)}{5}$

$= \frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{5}$

$\frac{1}{5}n^5- \frac{1}{2}n^4- \frac{1}{3}n^3- \frac{1}{30}n$

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