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← JEE Main 2022 · 25 Jul · Shift 1
MathematicsInteger+4 / -1Question 25 of 90

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 25

Let a1=b1=1a_1=b_1=1, an=an−1+2a_n=a_{n-1}+2 and bn=an+bn−1b_n=a_n+b_{n-1} for every natural number n≥2n\ge2. Then ∑n=115an⋅bn\displaystyle\sum_{n=1}^{15}a_n\cdot b_n is equal to ______.

Answer and worked solution

Official answer: 27560

Terms. an=2n−1a_n=2n-1, bn=n2b_n=n^2. ∑115(2n−1)n2=2⋅14400−1240=27560\sum_{1}^{15}(2n-1)n^2=2\cdot14400-1240=27560.

Answer: 27560