Questions: ∑(n=1 to 45)(4n+9)

Solution Steps

To evaluate the sum $\sum_{n=1}^{45}(4n+9)$, we can break it down into two separate sums: $\sum_{n=1}^{45}4n$ and $\sum_{n=1}^{45}9$. The first sum is an arithmetic series, and the second sum is simply 9 added 45 times. We can calculate each part separately and then add the results together.

We start by calculating the sum of $4n$ from $n=1$ to $n=45$: $$\sum_{n=1}^{45} 4n = 4 \sum_{n=1}^{45} n$$ The sum of the first $N$ natural numbers is given by the formula: $$\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$$ For $N=45$: $$\sum_{n=1}^{45} n = \frac{45 \cdot 46}{2} = 1035$$ Thus, $$\sum_{n=1}^{45} 4n = 4 \cdot 1035 = 4140$$

Next, we calculate the sum of the constant $9$ added $45$ times: $$\sum_{n=1}^{45} 9 = 9 \cdot 45 = 405$$

Now, we combine the results from the two sums: $$\sum_{n=1}^{45} (4n + 9) = \sum_{n=1}^{45} 4n + \sum_{n=1}^{45} 9 = 4140 + 405 = 4545$$

Final Answer