Question

求和.Sn=1*2+4*2平方+7*2的3次方+…+(3n-2)*2的n次方

Answer 1

∵(3n-2)*2^n=3n*2^n-2^(n+1)

∴原式=3*1*2-2^2+3*2*2^2-2^3+3*3*2^3-2^4+……+3*n*2^n-2^(n+1)

=3(1*2+2*2^2+3*2^3+4*2^4+……+n*2^n)-[2^2+2^3+2^4+……+2^(n+1)]

设S1=1*2+2*2^2+3*2^3+……+n*2^n

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