Question

If a sequence or series is not a direct form of an A.P, G.P etc., then its $n^{th}$ term cannot be determined. In such cases, we use the following steps to find the $n^{th}$ term $T_n$ of the given sequence.
Step $1:$ Find the differences between the successive terms of the given sequence.If these differences are in A.P, then take $T_n=a{n}^2+bn+c,$ where $a,b,c$ are constants.
Step $2:$ If the successive differences finding in step$1$ are in G.P with common ratio $r,$ then take $T_n=a+bn+c{r}^{n-1},$ where $a,b,c$ are constants.
Step $3:$ If the second successive differences (Differences of the differences) in step$1$ are in A.P, then take $T_n=a{n}^3+b{n}^2+cn+d,$ where $a,b,c,d$ are constants.
Step $4:$ If the second successive differences(Differences of the differences) in step$1$ are in G.P, then take $T_n=a{n}^2+bn+c+d{r}^{n-1},$ where $a,b,c,d$ are constants.
Now let sequences be$:$
$A:1,6,18,40,75,126,...$
On the basis of above information, answer the following questions:

If the $n^{th}$ term $T_n$ of the sequence $A$ is $a{n}^3+b{n}^2+cn+d,$ then $6a+2b-d$ is:

• A. $\ln 2$
• B. $2$
• C. $\ln 4$
• D. $4$

Answer 1

The correct option is D $4$

Given$:$ For a sequence $A:1,6,18,40,75,126,...$

First successive difference is $5,12,22,35,51,...$

Second successive difference is $7,10,13,16,...$

$\because T_n=a{n}^3+b{n}^2+cn+d$
For $n=1,$

$\therefore T_1=a+b+c+d=1$ ......$\left(1\right)$ ($T_1=1$ from the sequence $A$) and

For $n=2,$

$\therefore T_2=a{\left(2\right)}^3+b{\left(2\right)}^2+2c+d=\mathrm{6....}$(2) ($T_2=6$ from the sequence $A$)

Eqn $\left(2\right)-2\times$ Eqn $\left(1\right)$, we get
$T_2-2T_1=8a+4b+2c+d-2(a+b+c+d)=6-2\times 1$

$\Rightarrow T_2-2T_1=8a+4b+2c+d-2a-2b-2c-2d=6-2$
$\Rightarrow 6a+2b-d=4$