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#