JEE Main 2022 — Sequences And Series Question with Solution

Given,

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

$$\Rightarrow a_n-a_{n-1}=2$$

$$\therefore$$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.

Also given $$a_1=1$$

$$\therefore$$ Series is = 1, 3, 5, 7 ......

$$\therefore$$ $$a_n=1+(n-1)2=2n-1$$

Also $$b_n=a_n+b_{n-1}$$

When $$n=2$$ then

$$b_2-b_1=a_2=3$$

$$\Rightarrow b_2-1=3$$ [Given $$b_1=1$$]

$$\Rightarrow b_2=4$$

When $$n=3$$ then

$$b_3-b_2=a_3$$

$$\Rightarrow b_3-4=5$$

$$\Rightarrow b_3=9$$

$$\therefore$$ Series is = 1, 4, 9 ......

= 1², 2², 3² ....... n²

$$\therefore$$ $$b_n=n^2$$

Now, $$\sum _{n=1}^{15}\left(a_n.b_n\right)$$

$$=\sum _{n=1}^{15}\left[(2n-1)n^2\right]$$

$$=\sum _{n=1}^{15}2n^3-\sum _{n=1}^{15}{n}^2$$

$$=2\left(1^3+2^3+...{15}^3\right)-\left(1^2+2^2+...{15}^2\right)$$

$$=2\times {\left(\frac{15\times 16}{2}\right)}^2-\left(\frac{15(16)\times 31}{6}\right)$$

$$=27560$$