Question

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

Answer 1

The sequence of first consecutive differences is $3,7,19,55,....$.

The sequence of the second consecutive difference is $4,12,36,....$ .

Clearly, it is a GP with common ratio $3$.

Then, nth term of the given series be

$T_n=a(3{)}^{n-1}+b{n}^2+cn+d...............(i)$
Putting values of $n=1,2,3,4$ we get

$2=a+b+c+d.........\left(ii\right)$

$5=3a+4b+2c+d........\left(iii\right)$

$12=9a+9b+3c+d........\left(iv\right)$

$31=27a+16b+4c+d.......\left(v\right)$
After solving these equations, we get

$a=1,b=0,c=1,d=0$
a=1,b=0,c=1,d=0

Putting the values of $a,b,c,d$ in Eq. $\left(i\right)$, we get

$T_n=3^{n-1}+n$