If. Sn = m * 3 ^ n + 1, then the real number M

a1=s1=m*3+1=3m+1
a1+a2=s2=m*3^2+1
3m+1+a2=9m+1
a2=6m
a3=s3-s2=m*3^3-m*3^2=18m
Common ratio = A3 / A2 = 18m / 6m = 3
a2/a1=3
6m/(3m+1)=3
6m=3(3m+1)
m=-1

RELATED INFORMATIONS

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2. If the first n terms of the equal ratio sequence {an} and Sn = 3N + R, then r = () A. 0B. -1C. 1D. 3
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6. It is known that the sum of the first n terms of the sequence (an) is Sn, satisfying an + Sn = 2n. It is proved that the sequence (An-2) is an equal ratio sequence and an is obtained
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9. Let the first n terms of sequence {an} and Sn = 2an-2 ^ n (1) prove that {a (n + 1) - 2An} is the general term of finding {an} from the equal ratio sequence (2) The second question doesn't matter. Try to do it
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