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解答編― -129
1
=1/
= = n(n+1)(2n+1)+ 4·½ n(n+1) + 3n
2
= n(2n²+3n+1)+2n(n+1) +3n
=
1
= n{(2n² +3n+1)+(12n+12) +18}
6
=1/13m
-n(2n²+15n+31)
8
8
(5) (5x) = Σ k² – ≥ 2
k=1
k=1
=
8(8+1)(2+8+1)-8-2
_=(1+ener
=188 S
k=1
(6) (5¹)=2Σk+≥4-3Σk²
k=1
= 2.1/2.2
717-061 =
1/1/27(7+1)+7.4
S
k=1
6
-([ + IS-S)(1+IS) IS-—-—
1
[188=
-3.7(7+1)(2.7+1)
(8
(1) 12
S
