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 \(N = 44\).

Answer 2

Given:
\(\sum_{n=1}^N \left[ \frac{1}{5} + \frac{n}{25} \right] = 25\)

\(\sum_{n=1}^N \left[ \frac{5+n}{25} \right] = 25\)
For \(n= 1,2, 3, …….19\), the value of the function is \(0\).
For \(n= 20,21, 22, 23, …….44\), the value of the function is \(1\).
\(44=20+n-1\)
\(n= 44-20+1\)
\(n=25\) (Which is equal to the given value)
\(⇒ N = 44\)

So, the answer is \(44\).