求数列1,1-3,1-3+9,1-3+9-27,…前n项的和an = (-3)^0+ (-3)^1 +(-3)^2 +...+(-_数学解答

求数列1,1-3,1-3+9,1-3+9-27,…前n项的和
an = (-3)^0+ (-3)^1 +(-3)^2 +...+(-3)^(n-1)
= (1/4)[1- (-3)^n]
数列1,1-3,1-3+9,1-3+9-27,…前n项的和
这部分看不懂
=a1+a2+...+an
= (1/4) { n - (-3)[ 1- (-3)^n]/4 }
=(1/4) [ n + (3/4)(1-(-3)^n ) ]

人气：202 ℃　时间：2020-03-26 17:44:28

解答

=a1+a2+...+an
= (1/4){n-[(-3)+(-3)^2+(-3)^3+.+(-3)^n]}
=(1/4){n-(-3)[1-(-3)^n]/[1-(-3)]}
=(1/4){n+3*[1-(-3)^n]}/4
=(1/4)[n+3/4(1-(-3)^n)]