JEE Main 2022 — Sequences And Series Question with Solution

Given,

$$a_0=a_1=0$$

and $$a_{n+2}=3a_{n+1}-2a_n+1$$

For $$n=0,a_2=3a_1-2a_0+1$$

$$=3.0-2.0+1$$

$$=1$$

For $$n=1,a_3=3a_2-2a_1+1$$

$$=3.1-2.0+1$$

$$=4$$

For $$n=2,a_4=3a_3-2a_2+1$$

$$=3.4-2.1+1$$

$$=11$$

For $$n=3,a_5=3a_4-2a_3+1$$

$$=3.11-2.4+1$$

$$=26$$

For $$n=4,a_6=3a_5-2a_4+1$$

$$=3.26-2.11+1$$

$$=57$$

$$\therefore$$ $$S_n=1+4+11+26+57+....+t_n$$

$$S_n=1+4+11+26+....+t_{n-1}+t_n$$

$$0=1+3+7+15+31+.....-t_n$$

$$\Rightarrow t_n=1+3+7+15+31+....$$

Now, find the sum of the series,

$$t_n=1+3+7+15+31+.....+x_{n-1}+x_n$$ .....(1)

$$t_n=$$        $$1+3+7+15+.....+x_{n-1}+x_n$$ ......(2)

Subtracting (2) from (1), we get

-------------------------------------------------------------------------

$$0=1+2+4+8+16+....+x_n$$

$$\Rightarrow x_n=1+2+4+8+16+.....+$$ n terms

$$=\frac{1(2^n-1)}{2-1}$$

$$=2^n-1$$

$$\therefore$$ $$t_n=\sum _{n=1}^n{x}_n$$

$$=\sum _{n=1}^n(2^n-1)$$

$$=\sum _{n=1}^n{2}^n-\sum _{n=1}^{n}1$$

$$=\frac{2(2^n-1)}{2-1}-n$$

$$=2^{n+1}-2-n$$

$$t_1=2^2-2-1=1=a_2$$

$$t_2=2^3-2-2=4=a_3$$

$$t_3=2^4-2-3=11=a_4$$

$$\therefore$$ $$a_{22}=t_{21}=2^{22}-2-21=2^{22}-23$$

$$a_{23}=t_{22}=2^{23}-2-22=2^{23}-24$$

$$a_{24}=t_{23}=2^{24}-2-23=2^{24}-25$$

$$a_{25}=t_{24}=2^{25}-2-24=2^{25}-26$$

Now,

$$a_{25}{a}_{23}-2a_{25}{a}_{22}-2a_{23}{a}_{24}+4a_{22}{a}_{24}$$

$$=a_{25}(a_{23}-2a_{22})-2a_{24}(a_{23}-2a_{22})$$

$$=(a_{23}-2a_{22})(a_{25}-2a_{24})$$

$$=[(2^{23}-24)-2(2^{22}-23)][(2^{25}-26)-2(2^{24}-25)]$$

$$=[(2^{23}-24-2^{23}+46)][(2^{25}-26-2^{25}+50)]$$

$$=(22)(24)$$

$$=528$$