Question

Let a 1 = b 1 = 1, a n = a n – 1 + 2 and b n = a a + b n – 1 for every natural number _n _≥ 2. Then$\sum_{n=1}^{15}$ $a_n⋅b_n$ is equal to ______.

Answer 1

a 1 = b 1 = 1
a n = a n – 1 + 2 (for n ≥ 2) ; b n = a n + b n – 1
a 2 = a 1 + 2 = 1 + 2 = 3 ; b 2 = a 2 + b 1 = 3 + 1 = 4
a 3 = a 2 + 2 = 3 + 2 = 5 ; b 3 = a 3 + b 2 = 5 + 4 = 9
a 4 = a 3 + 2 = 5 + 2 = 7 ; b 4 = a 4 + b 3 = 7 + 9 = 16
a 15 = a 14 + 2 = 29
b 15 = 225

$\sum_{n=1}^{15}$ $a_nb_n$=1×1+3×4+5×9+⋯29×225

∴ $\sum_{n=1}^{11}$ $a_nb_n$=$\sum_{n=1}^{15}$(2n−1)n2=$\sum_{n=1}^{15}$ 2n3−$\sum_{n=1}^{15}$ n2

=2[$\frac{15×16}{2}$]2−[$\frac{15×16×31}{6}$]=27560