次の和を求めよ。
n
(1)k(k+1)
ずらす
k=1
gisid (= (k²+k)
=
=
k=1
n (n+1) (24+1) + n (n+1)
6
2
h (2n² +n+an+1)+3h33
(2) Zn-2-1
=
k=1
6
24 ² + 3 h² + 1 + 3 h² = 3
3
2
2n² + 6 h² +n+3
6
初項~項までの和
·α₁+ A2 + A3 +-+An
2
n
Zak
K=1
1 + 2 + 3 + n
1+2'
"
K=1
n
=
k
2
=
h(ith)
2
hch+1)
2
Ik² = n(n+1)(2n+1)
R=1
b
(3) 1-2+3-4+5.6++(2n-1)-2n
n
$(2R-1)2k
2
12
12:1
=
2
(4k²-2k)
R=
n(n+1)(2n+1)
n(n+1)
63
= 42 k² - 22 k = *^=
R=1
14 (25²+34 +1)-1-n
3
P
(1)(n+1Xn+2) (2) n(2"-1) (3) / n (n + 1x4n-1) = (n+D) (4n-
