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 $n \geq 2$.
Then $\displaystyle\sum_{n=1}^{15} a_n \cdot b_n$ is equal to ______

Answer 1

$ a _1= b _1=1$
$ a _2= a _1+2=3 $
$ a _3= a _2+2=5$
$ a _4= a _2+2=7$
$ \Rightarrow a _{ n }=2 n -1 $
$ b _2= a _1+ b _1=4 $
$ b _3= a _3+ b _2=9 $
$ b _4= a _4+ b _3=16$
$ b _{ n }= n ^2$
$ \displaystyle\sum_{n=1}^{15} a_n b_n $
$ \displaystyle\sum_{n=1}^{15}(2 n-1) n^2$
$ \displaystyle\sum_{n=1}^{15}\left(2 n^3-n^2\right)$
$=2 \frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2 n+1)}{6} $
Put $ n=15 $
$ =\frac{2 \times 225 \times 16 \times 16}{4}-\frac{15 \times 16 \times 31}{6}=27560$