(1) よって n k2+3kn+2n2 n (k+n)(k+2n) 2N n limΣ = lim Σ →k=n+1 k²+3kn+2n2 2錠 1 1 k+n k+2n 12 1 1 = lim - k=+1 k k +1 +2 n n 1 1 = k+n k+2n = *+1 12) dx=[log|x+1|-log|x+2] x+2 = log 3-log 4-(log 2-log 3)=2log 3-3log2

3 2020 電通大 2022 札医大, 2016 横国大 ] 2月 (1) 極限値 lim n を求めよ。 +12+3kn+2m2