数列的sn=n*2+n+1,则a8+a9+a10+a11+a12=?

问题描述:

数列的sn=n*2+n+1,则a8+a9+a10+a11+a12=?

S7 = a1 + a2 +a3 +.+a7 S12 = a1+a2 +a3 + .+a12 ==> a8+a9+a10+a11+a12= S12 - S7 = (12^2 + 12 +1) - (7^2 +7 +1) = (144 + 12 + 1) - (49 + 7 +1) = 157 - 57 = 100