Find the sum of the following series to `n` terms `5+7+13+31+85+`

Explanation

The sequence of first consecutive differences is `2,6,18,54,"…"`.Clearly, it is a GP with common ratio 3. Then, nth term of the given series be
`T_(n)=a(3)^(n-1)+bn+c " " "……(i)"`
Putting `n=1,2,3,` we get
`5=a+b+c " " "…..(ii)'`
`7=3a+2b+c " " "....(iii)"`
`13=9a+3b+c " " "....(iv)"`
Solving these equations, we get
`a=1,b=0,c=4`
Putting the values of a,b,c in Eq. (i), we get
`T_(n)=3^(n-1)+4`
Hence, sum of the series
`S_(n)=sumT_(n)=sum(3^(n-1)+4)=sum(3^(n-1))+4sum1`
`=(1+3+3^(2)+"..."+3^(n-1))+4n`
`=1*((3^(n-1)))/((3-1))+4n=(1)/(2)(3^(n)+8n-1)`