Progressions

Algebra Level 2

$\large \sum_{n=1}^{1729}n = \ ?$

4 solutions

Shivam Jadhav
Jun 13, 2015

$\boxed{\sum_{n=1}^{n}n=\frac{(n)(n+1)}{2}}$

Yes. Did using the same formula. Is there only one formula???

Hon Ming Rou - 5 years, 11 months ago

We can also apply the sum of series in A.P $\implies \large\frac{n}{2} \times (2a + (n - 1)d)$

Here, a = 1, d = 1, n = 1729 .

Another method is first to find the last term $l$

sum = $\frac{n}{2} \times (a + l)$ (here a is the first term)

Ram Mohith - 3 years ago

Francky Retice
Aug 3, 2015

h/2(a+l), where h= 1729 (the number of terms), a=1(the first term) and l = 1729(the last term). So, 1729/2(1+1729)= 864.5x1730= 1495585.

Lokesh Gurjar
Jun 14, 2015

By using sequence and series formula we get the answer

Yash Singh
Jun 13, 2015

simply apply the formula of the sum of first n natural numbers

n(n+1)/2

so is is 1729x1730/2 =1495585 calculator used obviously