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