The value of the `Sigma_(n=0)^(oo) (2n+3)/(3^n)` is equal to ______.

Answer

6

Explanation

We have S=`3+sum_(n=1)^(oo)(2n+3)/3^(n)=3+underset(S_(1))ubrace(sum_(n=1)^(oo)1/(3^(n-1)))+underset(S_(2))ubrace(sum_(n=1)^(oo)(2n)/(3^(n)))`
Now `S_(1)=sum_(n=1)^(oo)1/(3^(n-1))=1+1/3+1/3^(2)+…oo=1/(1-(1//3))=3/2`
`S_(2)=sum_(n=1)^(oo)(2n)/(3^(n))`
`thereforeS_(2)=2/3+4/3^(2)+6/3^(3)+8/3^(4)+....oo`
Now`((S_(2))/3=+2/3^(2)+4/3^(3)+6/3^(4)+..oo)/((2S_(2))/3=2/3[1+1/3+1/3^(2)+1/3^(3)+.....oo])` [On subtracting]
`thereforeS_(2)=1/(1-(1//3))=3/2`
Hence, `S=3+3/2+3/2=6`