Let a_1=b_1=1, a_n = a_n-1+2, and b_n = a_n+b_n-1 for every natural number n2.

Question

Let $a_1=b_1=1$, $a_n = a_{n-1}+2$, and $b_n = a_n+b_{n-1}$ for every natural number $n\ge2$. Then $\displaystyle\sum_{n=1}^{15}a_n\cdot b_n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 27560

Accepted Answer

$a_n = 2n-1$ (the odd numbers). $b_n = \sum_{k=1}^n a_k = n^2$ (sum of first $n$ odd numbers).

$$\sum_{n=1}^{15}a_nb_n = \sum_{n=1}^{15}(2n-1)n^2 = 2\sum n^3-\sum n^2 = 2(14400)-1240 = 27560.$$

Solutions · 1