Key Concepts and Formulas

• Arithmetic Progression (AP): The $n$-th term of an AP is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
• Summation Formulas:
  • Sum of first $N$ natural numbers: $\sum_{k=1}^{N} k = \frac{N(N+1)}{2}$
  • Sum of first $N$ squares: $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$
  • Sum of first $N$ cubes: $\sum_{k=1}^{N} k^3 = \left(\frac{N(N+1)}{2}\right)^2$
• Telescoping Sum: A sum of the form $\sum_{k=m}^{n} (f(k) - f(k-1))$ simplifies to $f(n) - f(m-1)$. This is useful for solving recurrence relations of the form $b_n - b_{n-1} = g(n)$.

Step-by-Step Solution

Step 1: Determine the general term $a_n$. We are given $a_1 = 1$ and $a_n = a_{n-1} + 2$ for $n \ge 2$. The relation $a_n - a_{n-1} = 2$ indicates that the sequence $a_n$ is an Arithmetic Progression with first term $a_1 = 1$ and common difference $d = 2$. Using the formula for the $n$-th term of an AP, $a_n = a_1 + (n-1)d$: $a_n = 1 + (n-1)2$ $a_n = 1 + 2n - 2$ $a_n = 2n - 1$ Thus, the general term for the sequence $a_n$ is $a_n = 2n - 1$.

Step 2: Determine the general term $b_n$. We are given $b_1 = 1$ and $b_n = a_n + b_{n-1}$ for $n \ge 2$. Rearranging the recurrence relation, we get $b_n - b_{n-1} = a_n$. To find $b_n$, we can sum this difference from $k=2$ to $n$: $\sum_{k=2}^{n} (b_k - b_{k-1}) = \sum_{k=2}^{n} a_k$ The left side is a telescoping sum: $(b_n - b_{n-1}) + (b_{n-1} - b_{n-2}) + \dots + (b_2 - b_1) = b_n - b_1$. So, $b_n - b_1 = \sum_{k=2}^{n} a_k$. Substitute $b_1 = 1$ and $a_k = 2k - 1$: $b_n - 1 = \sum_{k=2}^{n} (2k - 1)$ We can evaluate the sum $\sum_{k=2}^{n} (2k - 1)$ by splitting it: $\sum_{k=2}^{n} (2k - 1) = 2 \sum_{k=2}^{n} k - \sum_{k=2}^{n} 1$ The sum of $k$ from $2$ to $n$ is $\sum_{k=1}^{n} k - 1 = \frac{n(n+1)}{2} - 1$. The sum of $1$ from $2$ to $n$ is $(n-1) \times 1 = n-1$. So, $2 \left(\frac{n(n+1)}{2} - 1\right) - (n-1)$ $= n(n+1) - 2 - n + 1$ $= n^2 + n - 2 - n + 1$ $= n^2 - 1$ Alternatively, we can notice that $\sum_{k=2}^{n} (2k - 1)$ is the sum of an AP with first term $2(2)-1=3$, last term $2n-1$, and $n-1$ terms. The sum is $\frac{n-1}{2}(3 + 2n-1) = \frac{n-1}{2}(2n+2) = (n-1)(n+1) = n^2 - 1$. Substituting this back into the equation for $b_n$: $b_n - 1 = n^2 - 1$ $b_n = n^2$ Thus, the general term for the sequence $b_n$ is $b_n = n^2$.

Step 3: Calculate the sum $\sum_{n=1}^{15} a_n b_n$. We need to find the sum of the product of the general terms $a_n$ and $b_n$: $a_n b_n = (2n - 1)(n^2)$ $a_n b_n = 2n^3 - n^2$ Now, we compute the sum from $n=1$ to $15$: $\sum_{n=1}^{15} a_n b_n = \sum_{n=1}^{15} (2n^3 - n^2)$ Using the linearity of summation: $= 2 \sum_{n=1}^{15} n^3 - \sum_{n=1}^{15} n^2$ Now, we use the standard summation formulas for $N=15$: For the sum of cubes: $\sum_{n=1}^{15} n^3 = \left(\frac{15(15+1)}{2}\right)^2 = \left(\frac{15 \times 16}{2}\right)^2 = (15 \times 8)^2 = (120)^2 = 14400$ For the sum of squares: $\sum_{n=1}^{15} n^2 = \frac{15(15+1)(2 \times 15 + 1)}{6} = \frac{15 \times 16 \times 31}{6}$ $= \frac{240 \times 31}{6} = 40 \times 31 = 1240$ Substitute these values back into the sum expression: $2(14400) - 1240$ $= 28800 - 1240$ $= 27560$

Common Mistakes & Tips

• Incorrectly applying summation limits: Ensure that when using telescoping sums or other summation techniques, the starting and ending indices are handled correctly.
• Arithmetic errors: Calculations involving large numbers, especially multiplication and subtraction, are prone to errors. Double-check your arithmetic.
• Forgetting the initial term: When deriving $b_n$, make sure to account for $b_1$ correctly, especially when dealing with sums starting from $k=2$.

Summary

The problem requires finding the general terms of two sequences, $a_n$ and $b_n$, defined by recurrence relations. The sequence $a_n$ is identified as an arithmetic progression, leading to $a_n = 2n-1$. The sequence $b_n$ is found using a telescoping sum approach on its recurrence relation, yielding $b_n = n^2$. Finally, the sum $\sum_{n=1}^{15} a_n b_n$ is calculated by substituting the general terms and applying standard summation formulas for powers of natural numbers, resulting in a value of 27560.

The final answer is $\boxed{27560}$.