Question:

Let a1 = b1 = 1, an = an – 1 + 2 and bn = aa + bn – 1 for every natural number ≥ 2. Then\(\sum_{n=1}^{15}\) \(a_n⋅b_n\) is equal to ______.

Updated On: Apr 16, 2026
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Correct Answer: 27560

Solution and Explanation

a1 = b1 = 1
an = an – 1 + 2 (for n≥ 2) ; bn = an + bn – 1
a2 = a1 + 2 = 1 + 2 = 3 ; b2 = a2 + b1 = 3 + 1 = 4
a3 = a2 + 2 = 3 + 2 = 5 ; b3 = a3 + b2 = 5 + 4 = 9 
a4 = a3 + 2 = 5 + 2 = 7 ; b4 = a4 + b3 = 7 + 9 = 16
a15 = a14 + 2 = 29 
b15 = 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

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