Friday, 26 April 2019

discrete mathematics - Proving the sum of the first $n$ natural numbers by induction




I am currently studying proving by induction but I am faced with a problem.



I need to solve by induction the following question.



$$1+2+3+\ldots+n=\frac{1}{2}n(n+1)$$



for all $n > 1$.



Any help on how to solve this would be appreciated.







This is what I have done so far.



Show truth for $N = 1$



Left Hand Side = 1



Right Hand Side = $\frac{1}{2} (1) (1+1) = 1$




Suppose truth for $N = k$



$$1 + 2 + 3 + ... + k = \frac{1}{2} k(k+1)$$



Proof that the equation is true for $N = k + 1$



$$1 + 2 + 3 + ... + k + (k + 1)$$



Which is Equal To




$$\frac{1}{2} k (k + 1) + (k + 1)$$



This is where I'm stuck, I don't know what else to do. The answer should be:



$$\frac{1}{2} (k+1) (k+1+1)$$



Which is equal to:



$$\frac{1}{2} (k+1) (k+2)$$




Right?



By the way sorry about the formatting, I'm still new.


Answer



Basic algebra is what's causing the problems: you reached the point



$$\frac{1}{2}K\color{red}{(K+1)}+\color{red}{(K+1)}\;\;\;\:(**)$$



Now just factor out the red terms:




$$(**)\;\;\;=\color{red}{(K+1)}\left(\frac{1}{2}K+1\right)=\color{red}{(K+1)}\left(\frac{K+2}{2}\right)=\frac{1}{2}(K+1)(K+2)$$


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