Question 17334: Here is a small challenge for your entertainment and gratification. For any natural number n , let f(n) denote the sum of the numbers from 1 to n. Thus f(1)=1,f(2)=1+2=3 f(3)=1+2+3=6,f(x)=1+2+3+....+100=5050, etc.
It turns out that f is a polynomial of degree 2 in n. Figure out the coefficients of f:
f(n)=_?__n^2+_?__n+_?_,
There is a story about Carl Friedrich Gauss (1777-1855) who may have been the most outstanding mathematician in human history. According to the story, when Gauss was seven years old, his teacher at one stage was unhappy with the class and as a punishment he asked them to compute f(100) . Gauss' class mates started writing the numbers from 1 to 100 on their paper, and adding those numbers. Gauss stared at the ceiling and then wrote the single number 5050 on the sheet and handed it in. You aren't Gauss, but you also aren't seven years old, so maybe you can figure out what he was thinking!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! this is the sum of n consecutive natural numbers from 1 to n .they are in arithmatic progression called a.p the derivation for sum of a.p is done by this simple method of inversion
say S(100)=1+2+3......+98+99+100....reverse this order of summation
S(100)=100+99+98+....3+2+1
adding 2*S(100)=101+101+101+........101+101+101......100 times
2S(100)=101*100
S(100)=101*100/2=5050...is the method...similarly for n numbers we get
S(n)=1+2+3......+(n-2)+(n-1)+(n)....reverse this order of summation
S(n)=(n)+(n-1)+(n-2)+....3+2+1...adding
2*S(n)=(n+1)+(n+1)+(n+1)........(n+1)+(n+1)+(n+1).....n times
2S(n)=n(n+1)
S(n)=(n^2+n)/2=(1/2)*n^2+(1/2)*n+0
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