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

Answer 1

#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#

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Answer 2

Charles Anctil

#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#

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Answer 3

Ezekiel Alexander

To find the partial sum of the series Σn from n=1 to 50, you can use the formula for the sum of an arithmetic series, which is (n/2) * (first term + last term). In this case, the first term is 1 and the last term is 50. Therefore, the partial sum is (50/2) * (1 + 50) = 25 * 51 = 1275.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.