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8の(1)の答えがこうなりました
解答とは違うのですがこれで丸が付きますか?
8 公式を用いたΣの計算, Σが3重の和 [3TRIAL 数学B 問題58]
次の和を求めよ。
(1) 2 (4k³-1)
k=1
(2) Ź (2k-1)(2k+3)k
k=1
(解説
(1) (5)=4Σ k³ - Σ1=4_n(n+1)
(与式=4
= 4 { / {m^{n+1)] ²³-
21=4
k=1
k=1
=n(n+1)2-n
=n{n(n2+2n+1)-1}
=n(n3+2m²+n-1)
72
2
-n
71
(3) Z
m=1
h 4 + 2n
3
th-h
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わかりました!
ありがとうございます