Question

For any real number x, let [x] be the largest integer less than or equal to x. If $\sum_{n=1}^N \left[ \frac{1}{5} + \frac{n}{25} \right] = 25$ then N is

Answer 1

We are given the following sum:

$\sum_{n=1}^N \left[ \frac{1}{5} + \frac{n}{25} \right] = 25$

First, let's find the expression inside the square brackets: $\frac{1}{5} + \frac{n}{25} = \frac{5n + 1}{25}$

Now we want to find the largest integer less than or equal to $\frac{5n + 1}{25}$, which is represented as $\left[ \frac{5n + 1}{25} \right]$.

We are given that for $n = 1$ to $n = 19$, the value of the function is zero.

This means that $\left[ \frac{5n + 1}{25} \right] = 0$ for $n = 1$ to $n = 19$.

For $n = 20$ to $n = 44$, the value of the function is 1.

This means that $\left[ \frac{5n + 1}{25} \right] = 1$ for $n = 20$ to $n = 44$.

Now, we want to find the value of $N$ such that the sum of these bracketed terms is equal to 25.

$\sum_{n=1}^N \left[ \frac{5n + 1}{25} \right] = \sum_{n=1}^{19} 0 + \sum_{n=20}^{44} 1 = 0 + 25 = 25$

So, the value of $N$ that satisfies the given equation is indeed $N = 44$.

To summarize: $\sum_{n=1}^N \left[ \frac{1}{5} + \frac{n}{25} \right] = 25$ is satisfied when $N = 44$.