Find the sum a+(a+1)+(a+2)+…….(a+n-1) in terms of a and n
no way any factorials could be involved in such a sum
There are n terms, so we have
na + 0+1+2+...+(n-1)
We all know the sum of the first n-1 integers is (n-1)n/2, so the final sum is
na + n(n-1)/2 = n(a+n-1)/2
n!*a + (n-1)!
To find the sum of the arithmetic series a + (a + 1) + (a + 2) + ... + (a + n - 1) in terms of a and n, we can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is given by:
Sn = n/2 * (a1 + an)
where Sn is the sum of the series, n is the number of terms, a1 is the first term, and an is the last term.
In this case, the first term (a1) is equal to a, and the last term (an) can be found by adding n - 1 to the first term. So the last term is (a + n - 1).
Now we can substitute these values into the formula:
Sn = n/2 * (a + a + n - 1)
Simplifying the expression inside the parentheses:
Sn = n/2 * (2a + n - 1)
Further simplifying by distributing n/2 to the terms inside the parentheses:
Sn = n/2 * 2a + n/2 * (n - 1)
Simplifying:
Sn = n * a + (n * (n - 1))/2
Thus, the sum of the series a + (a + 1) + (a + 2) + ... + (a + n - 1) in terms of a and n is n * a + (n * (n - 1))/2.