设等比数列an的前n项和为Sn=4^n+m,则常数m=

设等比数列an的前n项和为Sn=4^n+m,则常数m=
设等比数列an的前n项和为Sn=4^n+m,则常数m=

设等比数列an的前n项和为Sn=4^n+m,则常数m=
Sn=(a1-anq)/(1-q)=4^n+m 化得a1-a1*q^n=(1-q)*m+(1-q)*4^n 得出a1=q-1 所以m=-1