SOLUTION: find the sum of the integers 1to50 inclusive

Algebra ->  Probability-and-statistics -> SOLUTION: find the sum of the integers 1to50 inclusive      Log On


   

Question 568361: find the sum of the integers 1to50 inclusive
Found 3 solutions by Edwin McCravy, stanbon, htmentor:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Let the sum be S

 S =  1 +  2 +  3 + ... + 48 + 49 + 50   <-- there are 50 terms
 S = 50 + 49 + 48 + ... +  3 +  2 +  1   <-- there are 50 terms
  
add the two equations term by term:

 S =  1 +  2 +  3 + ... + 48 + 49 + 50   <-- there are 50 terms
 S = 50 + 49 + 48 + ... +  3 +  2 +  1   <-- there are 50 terms
2S = 51 + 51 + 51 + ... + 51 + 51 + 51   <-- there are 50 terms
2S = 51×50  <--51 times 50 because there are 50 51's added together                           
2S = 2550
 S = 1275 

Or by formula: n%28n%2B1%29%2F2

50%2850%2B1%29%2F2 = 1275

Edwin

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find the sum of the integers 1to50 inclusive
---
S(50) = (50/2)(51)
----
S(50) = 25*51 = 1275
====
Cheers,
Stan H.
===========

Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
find the sum of the integers 1to50 inclusive
=====================================
This is an arithmetic series with a_1 = 1, a_50 = 50
The sum of an arithmetic series is
S_n = (n/2)*(a_1 + a_n)
Substituting the values, we get
S_50 = (50/2)*(1+50) = 1275