47 above sigma k=1 under sigma, k beside sigma

Question

watchmath · Answer

47*48/2

anonymous · Answer

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anonymous · Answer

47$$\sum_{k=1}^{47}k$$

anonymous · Answer

okay that is K(K+1)/2, as watch math pointed out, the answer is 47*48/2

anonymous · Answer

THANK YOU!

anonymous · Answer

in  general
$$\sum_{1}^{n} k = n(n+1)/2$$

anonymous · Answer

i forgot that formula, thanks for the generalization

anonymous · Answer

you are welcome

mathteacher1729 · Answer

$$\sum_{k=1}^{47}k$$ 

This is asking you to add up all the numbers from 1 to 47. 

There is a really nice visual proof of this attached.

Basically to find the sum of the first n integers multiply (n) times (n+1) and divide that 1/2. 

VISUALLY this is just finding the area of a rectangle with side lengths (n) and (n+1).  (Make sure you can see this in the picture, it'll be awesome when it clicks!) :)

anonymous · Answer

that is really helpful, thanks for ur help!

mathteacher1729 · Answer

... correction. 

"multiply (n) times (n+1) and divide that by TWO" 



(dividing by 1/2 mens multiplying by two... silly me, I was typing too fast in my excitement!)