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

Explanation

The sequence of difference between successive terms is 2,6,18,54,.. Clearly, it is a G.P. Let `T_(n)` be the nth term of the given series and `S_(n)` be the sum of its n terms. Then,
`{:(S_(n)=5+7+13+31+..+T_(n-1)+T_(n)" "(1)),(S_(n)=" "5+7+13+31+..+T_(n-1)+T_(n)" "(2)),(bar(0=5+[2+6+18+..+(n-1)" terms"]-T_(n))):}`
[Subtracting (2) from (1)]
or `0=5+2((3^(n-1)-1))/((3-1))-T_(n)`
or `T_(n)=5+(3^(n-1)-1)=4+3^(n-1)`
`thereforeS_(n)=sum_(k=1)^(n)T_(k)`
`=sum_(k=1)^(n)(4+3^(k-1))`
`=sum_(k=1)^(n)4+sum_(k=1)^(n)3^(k-1)`
`=4n+(1+3+3^(2)+...+3^(n-1))`
`=4n+1xx((3^(n)-1)/(3-1))`
`=4n+((3^(n)-1)/2)`
`=1/2[3^(n)+8n-1]`
Alternative method:
When the differences of consecutive terms are in G.P., nth term is given by, `T_(n)=ar^(n)+b`, where r is common ratio of the G.P.. Here, common ratio of G.P. obtained by differences is,r=3.
So, `T_(n)=a(3)^(n)+b`
`T_(1)=3a+b=5 ...(1)`
`T_(2)=9a+b=7 ...(2)`
Solving, we get a=1/3,b=4
So, `T_(n)=3^(n-1)+4`
`thereforeS_(n)=sum_(k=1)^(n)T_(k)=1/2(3^(n)+8n-1)`