数学のブログ

“離散的”な世界 数列とその和 平方の和、立方の和 和の級数、展開

新装版 数学読本3 (松坂 和夫(著)、岩波書店)の第13章(“離散的”な世界 - 数列)、13.1(数列とその和)、平方の和、立方の和の問20の解答を求めてみる。

1

∑ k = 1 n ( 3 k - 2 )
= 3 ∑ k = 1 n k - 2 n
= 3 n ( n + 1 ) 2 - 2 n
= 1 2 n ( 3 n - 1 )

2

∑ i = 1 n ( i + 1 ) ( 2 i - 1 )
= ∑ i = 1 n ( 2 i 2 + i - 1 )
= 2 · n ( n + 1 ) ( 2 n + 1 ) 6 + n ( n + 1 ) 2 - n
= 1 6 n ( 2 ( n + 1 ) ( 2 n + 1 ) + 3 ( n + 1 ) - 6 )
= 1 6 n ( ( n + 1 ) ( 4 n + 5 ) - 6 )
= 1 6 n ( 4 n 2 + 9 n - 1 )

3

∑ i = 1 n i ( i + 2 )
= ∑ i = 1 n i 2 + 2 ∑ i = 1 n i
= 1 6 n ( n + 1 ) ( 2 n + 1 ) + n ( n + 1 )
= 1 6 n ( n + 1 ) ( 2 n + 7 )

4

∑ j = 1 n j 2 + 3 ∑ j = 1 n j - 4 n
= 1 6 n ( n + 1 ) ( 2 n + 1 ) + 3 2 n ( n + 1 ) - 4 n
= 1 6 n ( ( n + 1 ) ( 2 n + 1 ) + 9 ( n + 1 ) - 24 )
= 1 6 n ( ( n + 1 ) ( 2 n + 10 ) - 24 )
= 1 3 n ( ( n + 1 ) ( n + 5 ) - 12 )
= 1 3 n ( n 2 + 6 n - 7 )
= 1 3 n ( n + 7 ) ( n - 1 )

5

∑ k = 1 n k ( k + 1 ) 2
= ∑ k = 1 n k ( k 2 + 2 k + 1 )
= ∑ k = 1 n k 3 + 2 ∑ k = 1 n k 2 + ∑ k = 1 n k
= ( 1 2 n ( n + 1 ) ) 2 + 1 3 n ( n + 1 ) ( 2 n + 1 ) + 1 2 n ( n + 1 )
= 1 12 n ( n + 1 ) ( 3 n ( n + 1 ) + 4 ( 2 n + 1 ) + 6 )
= 1 12 n ( n + 1 ) ( 3 n 2 + 11 n + 10 )
= 1 12 n ( n + 1 ) ( n + 2 ) ( 3 n + 5 )

6

∑ k = 1 n ( 2 k 3 + 3 k 2 + k )
= 1 2 n 2 ( n + 1 ) 2 + 1 2 n ( n + 1 ) ( 2 n + 1 ) + 1 2 n ( n + 1 )
= 1 2 n ( n + 1 ) ( n ( n + 1 ) + 2 n + 1 + 1 )
= 1 2 n ( n + 1 ) ( n ( n + 1 ) + 2 ( n + 1 ) )
= 1 2 n ( n + 1 ) 2 ( n + 2 )

コード(Wolfram Language, Jupyter)

an[k_] := 3k-2

Sum[an[k], {k, 1, n}]
Output
Factor[%]
Output
TraditionalForm[%]
Output
ListPlot[Table[an[k], {k, 1, 10}]]
Output
Sum[(i+1)(2i-1), {i, 1, n}]
Output
Sum[i(i+2), {i, 1, n}]
Output
Sum[(j^2+3j-4), {j, 1, n}]
Output
Factor[%]
Output
% // TraditionalForm
Output
Sum[k(k+1)^2, {k, 1, n}]
Output
Sum[2k^3+3k^2+k, {k, 1, n}]
Output