JEE 2022 — Mathematics PYQ

1. Finding the General Term $a_n$

The given recurrence relation is $a_{n+2}-2a_{n+1}+a_n=1$.

Notice that $a_{n+2}-2a_{n+1}+a_n$ is the second-order difference. Let's calculate the first few terms:

• $a_0=0$

• $a_1=0$

• For $n=0:a_2=2a_1-a_0+1=2(0)-0+1=1$

• For $n=1:a_3=2a_2-a_1+1=2(1)-0+1=3$

• For $n=2:a_4=2a_3-a_2+1=2(3)-1+1=6$

• For $n=3:a_5=2a_4-a_3+1=2(6)-3+1=10$

The sequence is $0,0,1,3,6,10,\dots$ which are triangular numbers shifted. The general formula for these terms is:

2. Calculating the Sum

We need to find $S={\sum }_{n=2}^{\infty }\frac{a_n}{7^n}$. Substituting $a_n$:

Let $x=\frac{1}{7}$. The sum becomes $\frac{1}{2}{\sum }_{n=2}^{\infty }n(n-1)x^n$.

Recall the power series for $(1-x{)}^{-1}={\sum }_{n=0}^{\infty }{x}^n$.

Differentiating twice with respect to $x$:

1. $\frac{d}{dx}(1-x{)}^{-1}=(1-x{)}^{-2}={\sum }_{n=1}^{\infty }n{x}^{n-1}$

2. $\frac{d^2}{d{x}^2}(1-x{)}^{-1}=2(1-x{)}^{-3}={\sum }_{n=2}^{\infty }n(n-1)x^{n-2}$

To get $n(n-1)x^n$, we multiply by $x^2$:

3. Final Computation

Substitute $x=\frac{1}{7}$ into the formula:

Correct Option: (B)