Монгол Бодлогын Сан

Эх хэлээрээ суралцаж, эх хэлээрээ мэдлэгээ түгээе.

Дараах нийлбэрүүдийг ол.

Бодлогын төрөл: Уламжлалт

Заавар:

Бодолт:

\begin{align*} \sum\limits_{k=3}^{21}(k-1)(k+3)&=\sum\limits_{k=1}^{21}(k^2+2k-3)-\sum\limits_{k=1}^{2}(k^2+2k-3)\\ &=\sum\limits_{k=1}^{21}k^2+2\sum\limits_{k=1}^{21}k-\sum\limits_{k=1}^{21}3-(1^2+2\cdot 1-3)-(2^2+2\cdot 2-3)\\ &=\dfrac{21\cdot(21+1)(2\cdot21+1)}{6}+2\cdot\dfrac{21(21+1)}{2}-3\cdot 21-0-5\\ &=3705 \end{align*}
$k(k+1)(k+2)=\dfrac{k(k+1)(k+2)(k+3)}{4}-\dfrac{(k-1)k(k+1)(k+2)}{4}$ тул \begin{align*} \sum\limits_{k=0}^n k(k+1)(k+2)&=\sum\limits_{k=0}^n\left(\dfrac{k(k+1)(k+2)(k+3)}{4}-\dfrac{(k-1)k(k+1)(k+2)}{4}\right)\\ &=\sum\limits_{k=0}^n\dfrac{k(k+1)(k+2)(k+3)}{4}-\sum\limits_{k=0}^n\dfrac{(k-1)k(k+1)(k+2)}{4}\\ &=\sum\limits_{k=1}^n\dfrac{k(k+1)(k+2)(k+3)}{4}-\sum\limits_{k=2}^n\dfrac{(k-1)k(k+1)(k+2)}{4}\\ &=\sum\limits_{k=1}^n\dfrac{k(k+1)(k+2)(k+3)}{4}-\sum\limits_{k=1}^{n-1}\dfrac{k(k+1)(k+2)(k+3)}{4}\\ &=\dfrac{n(n+1)(n+2)(n+3)}{4}=\dfrac{n^4+6n^3+11n^2+6n}{4} \end{align*}
$\sum\limits_{k=1}^n(3^k+k^2-3)$;
$1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$.