Question

Find the $n^{\text{th}}$ term of the series $2+5+12+31+86+\cdots$

• A. $T_n=(n-1{)}^3+n$
• B. None of the above
• C. $T_n=3^{n-1}+2n^2+n$
• D. $T_n=3^{n-1}+n$

Answer 1

The correct option is D $T_n=3^{n-1}+n$

The sequence of first consecutive differences is $3,7,19,55\cdots$
The sequence of the second consecutive differences is $4,12,36,\cdots$
Clearly, it is a G.P. with common ratio $3.$
The $n^{\text{th}}$ term of the given series is
$$T_n=a(3{)}^{n-1}+b{n}^2+cn+d$$
Putting $n=1,2,3,4$, we get
$$\begin{array}{cc}2& =a+b+c+d\\ 5& =3a+4b+2c+d\\ 12& =9a+9b+3c+d\\ 31& =27a+16b+4c+d\end{array}$$
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
$T_n=3^{n-1}+n$