【数列】Σ（シグマ）の重要公式　５パターン【公式、計算方法、証明】

$$\sum_{k=1}^{n}(k^2+k)$$

$$=\sum_{k=1}^{n}k^2+\sum_{k=1}^{n}k$$

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

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

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

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

$$\sum_{k=1}^{n}(3k^2-k+4)$$

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

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

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

$$=\frac{1}{2}n\{(n+1)(2n+1)-(n+1)+8\}$$

$$=\frac{1}{2}n\{2n^2+2n+8\}$$

$$=n\{n^2+n+4\}$$

$$\sum_{k=1}^{n}\{nk+(\frac{1}{2})^k\}$$

$$=n\sum_{k=1}^{n}k+\sum_{k=1}^n(\frac{1}{2})^k$$

$\displaystyle{\sum_{k=1}^n(\frac{1}{2})^k}$は 初項$\dfrac{1}{2}$ 、公比$\dfrac{1}{2}$、 項数$n$の等比数列の和なので

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

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

$1\quad\sum_{k=1}^{n} a$の証明

$$\sum_{k=1}^{n} a=\underbrace{a+a+a+\cdots+a}_{n個}=na$$

$2\quad\sum_{k=1}^{n} k=\frac{1}{2}n(n+1)$の証明

$$\quad\sum_{k=1}^{n} k=\underbrace{1+2+3+\cdots+(n-1)+n}_{n個}$$

これは初項１、末項nの等差数列なので$\require{cancel}$

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

$3\quad\sum_{k=1}^{n} k^2=\frac{1}{6}n(n+1)(2n+1)$の証明

$$(k+1)^3-k^3=3k^2+3k+1$$

の式に、$k=1,2,3,\cdots,n$をそれぞれ代入して

$$\begin{array}{ccc} k=1,&\cancel{2^3}-1^3&=&3\times1^2+3\times1+1 \\ k=2,&\cancel{3^3}-\cancel{2^3}&=&3\times2^2+3\times2+1\\ k=3,&\cancel{4^3}-\cancel{3^3}&=&3\times3^2+3\times3+1\\ k=4,&\cancel{5^3}-\cancel{4^3}&=&3\times4^2+3\times4+1\\\cdots&&\cdots&\\k=n,&(n+1)^3-\cancel{n^3}&=&3\times n^2+3\times n+1\end{array}$$

左辺と右辺それぞれ足すと

$$(n+1)^3-1^3=3\sum_{k=1}^{n}k^2+3\sum_{k=1}^{n}k+1\times n$$

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

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

$$3\sum_{k=1}^{n}k^2=\frac{1}{2}n\{(2n^2+6n+6)-3(n+1)-2\}$$

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

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

$$\sum_{k=1}^{n}k^2=\frac{1}{6}n(n+1)(2n+1)$$

$4\quad\sum_{k=1}^{n} k^3=\{\frac{1}{2}n(n+1)\}^2$の証明

$$(k+1)^4-k^4=4k^3+6k^2+4k+1$$

の式に、$k=1,2,3,\cdots,n$をそれぞれ代入して

$$\begin{array}{ccc} k=1,&\cancel{2^4}-1^4&=&4\times1^3+6\times1^2+4\times1+1 \\ k=2,&\cancel{3^4}-\cancel{2^4}&=&4\times2^3+6\times2^2+4\times2+1\\ k=3,&\cancel{4^4}-\cancel{3^4}&=&4\times3^3+6\times3^2+4\times3+1\\ k=4,&\cancel{5^4}-\cancel{4^4}&=&4\times4^3+6\times4^2+4\times4+1\\\cdots&&\cdots&\\k=n,&(n+1)^4-\cancel{n^4}&=&4\times n^3+6\times n^2+4\times n+1\end{array}$$

左辺と右辺それぞれ足すと

$$(n+1)^4-1^4=4\sum_{k=1}^{n}k^3+6\sum_{k=1}^{n}k^2+4\sum_{k=1}^{n}k+1\times n$$

$$(n+1)^4-1^4=4\sum_{k=1}^{n}k^3+6\frac{1}{6}n(n+1)(2n+1)+4\frac{1}{2}n(n+1)+n$$

$\sum_{k=1}^{n}k^3$について整理すると

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

$5\quad\sum_{k=1}^{n} ar^{n-1}=\frac{a(r^n-1)}{r-1}$の証明

（左辺）$=\sum_{k=1}^{n} ar^{n-1}$

$$=a+ar^1+ar^2+ar^3+\cdots+ar^{n-1}$$

これは初項a、公比r、項数nの等差数列の和

$$S=a+ar^1+ar^2+ar^3+\cdots+ar^{n-1}$$

として

$$\begin{array}{c|cc} & S &=&a+&ar+ar^2+ar^3+\cdots+ar^{n-1}& \\ – & rS &=&&ar+ar^2+ar^3+\cdots+ar^{n-1}&+ar^{n}\\ \hline&S(1-r)&=&a&&-ar^{n} \end{array}$$

$$(1-r)S=a(1-r^{n})$$

よって

$$S=\frac{a(1-r^n)}{1-r}$$