Find the nth term of the series `2+5+12+31+86+"..."`.

Explanation

The sequence of first consecutive differences is `3,7,19,55,"…"`.The sequence of the second consecutive differences is `4,12,36,"..."`. Clearly, it is a GP with common ratio3. Then, nth term of the given series be
`T_(n)=a(3)^(n-1)+bn^(2)+cn+d"......(i)"`
Putting `n=1,2,3,4` we get
`2=a+b+c+d " " "…..(ii)'`
`5=3a+4b+2c+d" " "....(iii)"`
`12=9a+9b+3c+d" " ".....(iv)"`
`31=27a+16b+4c+d" " ".....(v)"`
After, solving these equations, we get
`a=1,b=0,c=1,d=0`
Putting the values of a,b,c,d in Eq. (i), we get
`T_(n)=3^(n-1)+n`