Question

Let \( a_{1}=b_{1}=1, a_{n}=a_{n-1}+2 \) and \( b_{n}=b_{n}+b_{n-1} \) for every natural number \( n \geq 2 \). Then \( \sum a_{a} b_{n} \) where n belong to 1 to 15 is

Answer 1

Correct answer: 27560

a₁ = b₁ = 1

a_n = a_{n – 1} + 2 (for n ≥ 2) ; b_n = a_n + b_{n – 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

\(\sum\limits_{n = 1}^{15} a_n b_n = 1\times 1+3\times 4+5\times 9+....29\times 225\)

\(\therefore \sum\limits_{n = 1}^{11} a_n b_n = \sum\limits_{n = 1}^{15} (2n -1)n^2 = \sum \limits_{n = 1}^{15}2n^3 - \sum\limits_{n=1}^{15}n^2\)

\(= 2\left[ \frac{15 \times 16}2\right]^2 -\left[\frac{15 \times 16 \times 31}6\right]\)

\(= 27560\)