The sequence first consists of all the starting from 1 to n and then remaining even starting up to n. Let’s suppose we have n as 00. Then the sequence becomes 1 3 5 7….999 4 6….1000

We are given a (L, R), we need to find sum of numbers of this sequence in given .

Note: Here the range is given as (L, R) L and R are included in the rangee

Examples:

```Input  : n = 10
Range 1 6
Output : 27
Explanation:
Sequence is 1 3 5 7 9 2 4 6 8 10
Sum in range (2, 6)
= 1 + 3 + 5 + 7 + 9 + 2
= 27

Input  : n = 5
Range 1 2
Output : 4
Explanation:
sequence is 1 3 5 2 4
sum = 1 + 3 = 4
```

The idea is to first find sum of numbers before left(excluding left), then find sum of numbers before right (including right). We get result as second sum minus first sum.

How to find sum till a limit?
We first count how many odd numbers are there, then we use formulas for sum of odd natural numbers and sum of even natural numbers to find the result.

How to find count of odd numbers?

• If n is odd then the number of odd numbers are ((n/2) + 1)
• If n is even then number of odd numbers are (n/2)

By simple observation, we get the number of odd numbers is ceil(n/2). So, the number of even numbers are n – ceil(n/2).

• Sum of first N odd numbers is (N^2)
• Sum of first N even numbers is (N^2) + N

For a given number x how will we find the sum in the sequence from 1 to x?
let’s suppose x is less than the number of odd numbers.

• Then we simply return (x*x)

If the x is greater then the number of odd numbers

• var = x-odd;
• That means we need first var even numbers
• we return (odd*odd) + (var*var) + var;
• ```// CPP program to find sum in the given range in
// the sequence 1 3 5 7.....N 2 4 6...N-1
#include <bits/stdc++.h>
using namespace std;

// For our convenience
#define ll long long

// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....N 2 4 6...N-1
ll sumTillX(ll x, ll n)
{
// number of odd numbers
ll odd = ceil(n / 2.0);

if (x <= odd)
return x * x;

// number of extra even
// numbers required
ll even = x - odd;

return ((odd * odd) + (even * even) + even);
}

int rangeSum(int N, int L, int R)
{
return sumTillX(R, N) - sumTillX(L-1, N);
}

// Driver code
int main()
{
ll N = 10, L = 1, R = 6;
cout << rangeSum(N, L, R);
return 0;
}
```

There is a difference between I CAN and CAN I

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