Question

na_po

Σの計算で黄色い部分がなぜこうなるのか至急教えてほしいです！よろしくお願いします！！

k=1
(4) k³=
24
k=1
=n(n
||
n
2
k=1
= n(n+1
= 1/12
n
1612
56 (1) (2k+3)=22k+23
k=1
・1/2n(n+1)+3n=n(n+1)+3ml - 1
k=1
・35(35+1)(2.35+1)=14910
= 2.
= n{(n+1)+3) = n(n+4)
n
(2) Σ (k² + k) = Σk²+Zk
k=1
k=1
k=1
1
-n(n+1)(2n+1)+ ½ n(n+1)
n
= -n(n+1)(n+2)
• 24(24+1)}² =
k=1
.
-n(n+1){(2n+1)+3) =
=
= n(2n² — 15n+13) =
+
(0 2
nes=
(3) Σ (k²−6k+5)= Σk² −6Σk+25
k=1
= n(n+1)(2n+1) −6 · ½-\n(n+1)+5n_J<£
を代入しそ
= n{(n+1)(2n + 1) − 18(n +1) +30)
yn
n
n
(4) Σ (k³-4k)= Σ k³ −4Σ k
k=1
k=1
k=1
=90000
n
-4
1
1
= n²(n+1) ²—4 · ½ n(n+1)
= n(n+1){n(n+1)−8}
-n(n+1)(n²+n-8)
n
=Σk²-Σk-Σ²
k=1
n
k=1
n
n
k=1
(5) Σ(k+1)(k-2)=(k²-k-2)
k=1
n(n+1)(2n+4)
TR
3+ [10) 8.8
+S+1 S)
(3) +8
1+1+1=12 (1) №
(n+1)
-n(n − 1)(2n — 13)
n(n
8-1-18-
= n(n+1X2
-
-n(n+1)(2n+1) n(n+1)-2nG
= n(n+1X2n+1)-3(n+1)-12)
1)(2n-1)
= n(n − 1){(2n – 1
=n(n-1)(2n-1
(n-1)(n-8)
||
=n(n-1
k
57 (1) (-3)*
18
1=6
+18-1-
18
(2) 212²=21²-2
・18(18+1)(2.1
n-
=
=1+(-1/2)+
1.{1-(- -1/3
1-(-133
n+1
k=0
=2109-55=2054
k=1
n
k=0
(3) (4k³-1)=4>
ARC=
=4 (n+1){(n+1
=(n+1){(n+1)(n −
=(n+1){(n+1)(n²
=(n+1)(n³+5n²+
(4) Σ(3k-1)² = (3.
3
n
5
= 1+ Σ(9k²-6k+
k=1
n+
n
=1+92 k²-62
k=1
k=1
=1+9=n(n+1)=
n(n+1
in(n+1)(2n+1)
(n+1){3n(2n+

□56 (1)
(2k+3)
n
*(4) Σ (k³-4k)
k=1
k=1
k=1
(2) Ź (k² + k)
k=1
k=1
n
(5) Σ (k+1)(k-2)
k=1
*(3)
*(6)
k=1
n-1
k=1
(k²-6k+5)
(k²-5k)
例題

σの計算 和の記号σ

Answer 1

カルボキシ基

こんな感じでわかりますか？

na_po

わかりました！
教えていただきありがとうございます🙇‍♀️