Let $a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$ and $b_{n}=a_{n}+b_{n-1}$ for every

natural number $n \geqslant 2$. Then $\sum\limits_{n = 1}^{15} {{a_n}.{b_n}}$ is equal to ___________.

<p>Given,</p> <p>${a_n} = {a_{n - 1}} + 2$</p> <p>$\Rightarrow {a_n} - {a_{n - 1}} = 2$</p> <p>$\therefore$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.</p> <p>Also given ${a_1} = 1$</p> <p>$\therefore$ Series is = 1, 3, 5, 7 ......</p> <p>$\therefore$ ${a_n} = 1 + (n - 1)2 = 2n - 1$</p> <p>Also ${b_n} = {a_n} + {b_{n - 1}}$</p> <p>When $n = 2$ then</p> <p>${b_2} - {b_1} = {a_2} = 3$</p> <p>$\Rightarrow {b_2} - 1 = 3$ [Given ${b_1} = 1$]</p> <p>$\Rightarrow {b_2} = 4$</p> <p>When $n = 3$ then</p> <p>${b_3} - {b_2} = {a_3}$</p> <p>$\Rightarrow {b_3} - 4 = 5$</p> <p>$\Rightarrow {b_3} = 9$</p> <p>$\therefore$ Series is = 1, 4, 9 ......</p> <p>= 1², 2², 3² ....... n²</p> <p>$\therefore$ ${b_n} = {n^2}$</p> <p>Now, $\sum\limits_{n = 1}^{15} {\left( {{a_n}\,.\,{b_n}} \right)}$</p> <p>$= \sum\limits_{n = 1}^{15} {\left[ {(2n - 1){n^2}} \right]}$</p> <p>$= \sum\limits_{n = 1}^{15} {2{n^3} - \sum\limits_{n = 1}^{15} {{n^2}} }$</p> <p>$$ = 2\left( {{1^3} + {2^3} + \,\,...\,\,{{15}^3}} \right) - \left( {{1^2} + {2^2} + \,\,...\,\,{{15}^2}} \right)$$</p> <p>$$ = 2 \times {\left( {{{15 \times 16} \over 2}} \right)^2} - \left( {{{15(16) \times 31} \over 6}} \right)$$</p> <p>$= 27560$</p>