Zadania na piątek 08 maja 2009

Zadanie 1.

Oblicz:

(1)
\begin{equation} 1^3+2^3+3^3+...+n^3=X \end{equation}

Rozwiązanie:
Wiemy, że:

(2)
\begin{align} \left\{\begin{array}{l l} 1^4=1\\ 2^4=(1+1)^4=1^4+4\cdot 1^3+6\cdot 1^2+4\cdot 1+1\\ 3^4=(2+1)^4=2^4+4\cdot 2^3+6\cdot 2^2+4\cdot 2+1\\ 4^4=(3+1)^4=3^4+4\cdot 3^3+6\cdot 3^2+4\cdot 3+1\\ \vdots\\ (n+1)^4=n^4+4\cdot n^3+6\cdot n^2+4\cdot n+1 \end{array}\right. \end{align}

Dodając stronami otrzymujemy:

(3)
\begin{equation} 1^4+2^4+3^4+...+n^4+(n+1)^4=(1^4+2^4+3^4+...+n^4)+4(1^3+2^3+3^3+...+n^3)+6(1^2+2^2+3^2+...+n^2)+4(1+2+3+...+n)+(n+1) \end{equation}

Podstawiając za $1^2+2^2+3^2+...+n^2$ i $1+2+3+...+n$ odpowiednie wzory, a za $1^3+2^3+3^3+...+n^3$ X otrzymujemy:

(4)
\begin{align} (n+1)^4=4X+6\cdot\frac{n(n+1)(2n+1)}{6}+4\cdot \frac{n(n+1)}{2}+(n+1) \end{align}
(5)
\begin{align} 4X=(n+1)^4-6\cdot\frac{n(n+1)(2n+1)}{6}-4\cdot \frac{n(n+1)}{2}-(n+1) \end{align}
(6)
\begin{align} X=\frac{(n+1)}{4}\left((n+1)^3-n(2n+1)-2n-1\right) \end{align}
(7)
\begin{align} X=\frac{(n+1)}{4}\left((n+1)^3-n(2n+1)-(2n+1)\right) \end{align}
(8)
\begin{align} X=\frac{(n+1)}{4}\left((n+1)^3-(2n+1)(n+1)\right) \end{align}
(9)
\begin{align} X=\frac{(n+1)^2}{4}\left((n+1)^2-(2n+1)\right) \end{align}
(10)
\begin{align} X=\frac{(n+1)^2}{4}\left((n+1)^2-2(n+1)+1\right) \end{align}
(11)
\begin{align} X=\frac{(n+1)^2}{4}(n+1-1)^2 \end{align}
(12)
\begin{align} X=\frac{n^2(n+1)^2}{4} \end{align}
(13)
\begin{align} 1^3+2^3+3^3+...+n^3=\frac{n^2(n+1)^2}{4} \end{align}

Zadanie 2.

Oblicz:

(14)
\begin{equation} 1^4+2^4+3^4+...+n^4=X \end{equation}

Rozwiązanie:
Wiemy, że:

(15)
\begin{align} \left\{\begin{array}{l l} 1^5=1\\ 2^5=(1+1)^5=1^5+5\cdot 1^4+10\cdot 1^3+10\cdot 1^2+5\cdot 1+1\\ 3^5=(2+1)^5=2^5+5\cdot 2^4+10\cdot 2^3+10\cdot 2^2+5\cdot 2+1\\ 4^5=(3+1)^5=3^5+5\cdot 3^4+10\cdot 3^3+10\cdot 3^2+5\cdot 3+1\\ \vdots\\ (n+1)^5=n^5+5\cdot n^4+10\cdot n^3+10\cdot n^2+5\cdot n+1 \end{array}\right. \end{align}

Dodając stronami otrzymujemy:

(16)
\begin{equation} 1^5+2^5+3^5+...+n^5+(n+1)^5=(1^5+2^5+3^5+...+n^5)+5(1^4+2^4+3^4+...+n^4)+10(1^3+2^3+3^3+...+n^3)+10(1^2+2^2+3^2+...+n^2)+5(1+2+3+...+n)+(n+1) \end{equation}

Podstawiając za $1^3+2^3+3^3+...+n^3$, $1^2+2^2+3^2+...+n^2$ i $1+2+3+...+n$ odpowiednie wzory, a za $1^4+2^4+3^4+...+n^4$ X otrzymujemy:

(17)
\begin{align} (n+1)^5=5X+10\cdot \frac{n^2(n+1)^2}{4}+10\cdot \frac{n(n+1)(2n+1)}{6}+5\cdot \frac{n(n+1)}{2}+(n+1) \end{align}
(18)
\begin{align} 5X=(n+1)^5-10\cdot \frac{n^2(n+1)^2}{4}-10\cdot \frac{n(n+1)(2n+1)}{6}-5\cdot \frac{n(n+1)}{2}-(n+1) \end{align}
(19)
\begin{align} X=\frac{n+1}{5}\left((n+1)^4-\frac{5n^2(n+1)}{2}-\frac{5n(2n+1)}{3}-\frac{5n}{2}-1\right) \end{align}
(20)
\begin{align} X=\frac{n+1}{5}\cdot\frac{6(n+1)^4-3\cdot 5n^2(n+1)-2\cdot 5n(2n+1)-3\cdot 5n-6}{6} \end{align}
(21)
\begin{align} X=\frac{n+1}{30}\left(6(n+1)^4-15n^2(n+1)-10n(2n+1)-15n-6\right) \end{align}
(22)
\begin{align} X=\frac{n+1}{30}\left(5(n+1)^4-15n^2(n+1)-10n(2n+2)+10n-15n-5+(n+1)^4-1\right) \end{align}
(23)
\begin{align} X=\frac{n+1}{30}\left(5(n+1)^4-15n^2(n+1)-20n(n+1)-5(n+1)+(n+1)^4-1\right) \end{align}
(24)
\begin{align} X=\frac{n+1}{30}\left(5(n+1)\left((n+1)^3-3n^2-4n-1\right)+(n+1)^4-1\right) \end{align}
(25)
\begin{align} X=\frac{n+1}{30}\left(5(n+1)\left(n^3+3n^2+3n+1-3n^2-4n-1\right)+(n+1)^4-1\right) \end{align}
(26)
\begin{align} X=\frac{n+1}{30}\left(5(n+1)\left(n^3-n\right)+(n+1)^4-1\right) \end{align}
(27)
\begin{align} X=\frac{n+1}{30}\left(5n(n+1)\left(n^2-1\right)+(n+1)^4-1\right) \end{align}
(28)
\begin{align} X=\frac{n+1}{30}\left(5n(n-1)(n+1)^2+(n+1)^4-1\right) \end{align}
(29)
\begin{align} X=\frac{n+1}{30}\left((n+1)^2\left(5n(n-1)+(n+1)^2\right)-1\right) \end{align}
(30)
\begin{align} X=\frac{n+1}{30}\left((n+1)^2\left(5n^2-5n+n^2+2n+1\right)-1\right) \end{align}
(31)
\begin{align} X=\frac{n+1}{30}\left((n+1)^2\left(6n^2-3n+1\right)-1\right) \end{align}
(32)
\begin{align} X=\frac{n+1}{30}\left((n+1)^2\left(3n(n-1)+1\right)-1\right) \end{align}
(33)
\begin{align} X=\frac{n+1}{30}\left(3n(n-1)(n+1)^2+(n+1)^2-1\right) \end{align}
(34)
\begin{align} X=\frac{n+1}{30}\left(3n(n^2-1)(n+1)+n^2+2n+1-1\right) \end{align}
(35)
\begin{align} X=\frac{n(n+1)}{30}\left(3n^3+3n^2-3n-3+n+2\right) \end{align}
(36)
\begin{align} X=\frac{n(n+1)}{30}\left(3n^3+3n^2-2n-1\right) \end{align}
(37)
\begin{align} X=\frac{n(n+1)(3n^2(n+1)-2n-1)}{30} \end{align}
O ile nie zaznaczono inaczej, treść tej strony objęta jest licencją Creative Commons Attribution-ShareAlike 3.0 License