数学の基礎と公式-10

数学の基礎と公式

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10. 級数-1

(1)$\displaystyle{}$

$1+2+3+\cdots + n=\displaystyle{} \sum_{k=1}^n k=\dfrac{1}{2}n(n+1)$

$1^2+2^2+3^2+\cdots + n^2=\displaystyle{} \sum_{k=1}^n k^2=\dfrac{1}{6}n(n+1)(2n+1)$

$\displaystyle{ \sum_{k=1}^n k^3={{n^2\,\left(n+1\right)^2}\over{4}}}$

$\displaystyle{ \sum_{k=1}^n k^4={{n\,\left(n+1\right)\,\left(2\,n+1\right)\,\left(3\,n^2+3\,n-1 \right)}\over{30}}}$

$\displaystyle{ \sum_{k=1}^n k^5={{n^2\,\left(n+1\right)^2\,\left(2\,n^2+2\,n-1\right)}\over{12}}}$

$\displaystyle{ \sum_{k=1}^n k^6={{n\,\left(n+1\right)\,\left(2\,n+1\right)\,\left(3\,n^4+6\,n^3-3\, n+1\right)}\over{42}}}$

$\displaystyle{ \sum_{k=1}^n k^7={{n^2\,\left(n+1\right)^2\,\left(3\,n^4+6\,n^3-n^2-4\,n+2\right) }\over{24}}}$

$\displaystyle{ \sum_{k=1}^n k^8}$

$\displaystyle{={{n\,\left(n+1\right)\,\left(2\,n+1\right)\,\left(5\,n^6+15\,n^5+5 \,n^4-15\,n^3-n^2+9\,n-3\right)}\over{90}}}$

$\displaystyle{ \sum_{k=1}^n k^9}$

$\displaystyle{ ={{n^2\,\left(n+1\right)^2\,\left(n^2+n-1\right)\,\left(2\,n^4+4\,n^ 3-n^2-3\,n+3\right)}\over{20}}}$

$\displaystyle{ \sum_{k=1}^n k^{10}}$

$\displaystyle{={{n\,\left(n+1\right)\,\left(2\,n+1\right)\,\left(n^2+n-1\right)\, \left(3\,n^6+9\,n^5+2\,n^4-11\,n^3+3\,n^2+10\,n-5\right)}\over{66}}}$

(2)

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

$\displaystyle{ \sum_{k=1}^n  \left( \dfrac{ k - 1 }{ 2 } \right)+\dfrac{1}{16}=\dfrac{(2n-1)^2}{4}}$

$\displaystyle{ \sum_{k=1}^n (8k-1)+1=(2n-1)^2}$

$\displaystyle{ \sum_{k=1}^n \left( \dfrac{ k - 1 }{ 8 } \right)+\dfrac{ 1 }{ 64 }=\dfrac{ ( 2n-1)^ 2 }{ 64 }}$

$\displaystyle{ \sum_{k=1}^n \left \lbrace k( k + 1) \right \rbrace={{n\,\left(n+1\right)\,\left(n+2\right)}\over{3}}}$

$\displaystyle{ \sum_{k=1}^n \left \lbrace k( k + 1)(k+2) \right \rbrace={{n\,\left(n+1\right)\,\left(n+2\right)\,\left(n+3\right)}\over{4}}}$

$\displaystyle{ \sum_{k=1}^n \left \lbrace k( k + 1)(k+2)(k+3) \right \rbrace={{n\,\left(n+1\right)\,\left(n+2\right)\,\left(n+3\right)\,\left(n+ 4\right)}\over{5}}}$

$\displaystyle{ \sum_{k=1}^n \left \lbrace k( k + 1)(k+2)(k+3)(k+4) \right \rbrace}$

$\displaystyle{={{n\,\left(n+1\right)\,\left(n+2\right)\,\left(n+3\right)\,\left(n+
 4\right)\,\left(n+5\right)}\over{6}}}$

$\displaystyle{ \sum_{k=1}^n \left( k x^{( k - 1)} \right)={{x^{n}\,\left(n\,x-n-1\right)}\over{\left(x-1\right)^2}}+{{1 }\over{\left(x-1\right)^2}}}$

$\displaystyle{ \sum_{k=1}^n \left( k(k+1) x^{( k - 1)} \right)}$

$\displaystyle{ ={{x^{n}\,\left(n^2\,x^2+n\,x^2-2\,n^2\,x-4\,n\,x+n^2+3\,n+2\right)
 }\over{\left(x-1\right)^3}}-{{2}\over{\left(x-1\right)^3}}}$