To find first five terms of the given series a1=3,an=3a(n−1)+2 for all n>1 and hence find the corresponding series.

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Solution

We have, a1=3,an=3a(n−1)+2∀n>1
Substitute n=2 in an=3an−1+2
⇒a2=3a1+2=3(3)+2=11
for n=3; a3=3a2+2=3(11)+2=35
For n=4; a4=3a3+2=3(35)+2=107
For n=5; a5=3a4+2=3(107)+2=323
Hence, the first five terms of the sequence are 3,11,35,107 and 323.
Corresponding series is 3+11+35+107+323+...

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