Question

Let a₁ = b₁ = 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

Correct Answer: 27560

Let's determine a_n and b_n for n ≥ 2 based on the given recursion: 

• a₁ = 1,
• a_n = a_n-1 + 2 for n ≥ 2. This implies a_n = 2n - 1.

Now, let's compute b_n:

• b₁ = 1,
• b_n = a_n + b_n-1.

Calculate b_n by plugging in the values of a_n:

• b2 = a2 + b1 = 3 + 1 = 4,
• b3 = a3 + b2 = 5 + 4 = 9,
• Continue this pattern up to n = 15.

We need to calculate \(\sum_{n=1}^{15} a_n \cdot b_n\):

n	an	bn	an · bn
1	1	1	1
2	3	4	12
3	5	9	45
4	7	16	112
5	9	25	225
6	11	36	396
7	13	49	637
8	15	64	960
9	17	81	1377
10	19	100	1900
11	21	121	2541
12	23	144	3312
13	25	169	4225
14	27	196	5292
15	29	225	6525

Compute the sum of products:

1 + 12 + 45 + 112 + 225 + 396 + 637 + 960 + 1377 + 1900 + 2541 + 3312 + 4225 + 5292 + 6525 = 27560.

Hence, the final sum is 27560, which falls within the specified range (27560, 27560).