How do you find the partial sum of $Sigma n$ from n=1 to 50?

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Answer 1 · 2017-12-03T13:38:41

$sum_(n=1)^50n=1275$

We can evaluate the sum the tedious way by just adding continually, but we can be smart. There is a formula for adding all the numbers up to $k$ :

$sum_(n=1)^kn=(k(k+1))/2$

In this case, it becomes:
$sum_(n=1)^50n=(50(50+1))/2=(50*51)/2=1275$

Answer 2 · 2017-12-03T13:49:30

$Sigma_(n=1)^50 n=color(blue)(1275)$

$s=(Sigma_(n=1)^50 n) = overbrace(color(white)("x")1+ ... + 50)^(50" terms")$
$s=(Sigma_(n=1)^50 n) = underbrace(50+ ... + 1)_(50" terms")$

$2scolor(white)("xxxxxxxxxx")=underbrace(51+ ... +51)_(50" terms")$

$color(white)("xxxxxxxxxxxxxx")= 51 xx 50$

$rarr s= (Sigma_(n=1)^(50) n)=51 xx25 =1275$

How do you find the partial sum of Sigma nSigma n from n=1 to 50?