Proof with mathematic induction
- Thread starter katy_042
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lev888
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Ok, please post the first 2 steps of the proof. Then we'll work on the 3rd.
HallsofIvy
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"Proof by induction" involves
1) Prove it is true when n= 1. That's easy, the left side is just 1 and the right side is \(\displaystyle \left(\frac{1}{2}(1)(1+1)\right)^2= ((1/2)(1)(2))^2= 1^2= 1\).
2) Prove that if this is true for n= k then it is also true for n= k+1.
For n= k that is \(\displaystyle \sum_{i=1}^{k} i^3= \left(\frac{1}{2}k(k+1)\right)^2\)
and you want to use that to show that
\(\displaystyle \sum_{i=1}^{k+1} i^3= \left(\frac{1}{2}k(k+1)\right)^2\).
The reason so many "induction problems" involve sums is because
the sum to k+ 1 is just the sum to k plus one more term. Here that is
\(\displaystyle \sum_{i=1}^{k} i^3+ (k+1)^3= \left(\frac{1}{2}k(k+1)\right)^2+ (k+1)^3\).
1) Prove it is true when n= 1. That's easy, the left side is just 1 and the right side is \(\displaystyle \left(\frac{1}{2}(1)(1+1)\right)^2= ((1/2)(1)(2))^2= 1^2= 1\).
2) Prove that if this is true for n= k then it is also true for n= k+1.
For n= k that is \(\displaystyle \sum_{i=1}^{k} i^3= \left(\frac{1}{2}k(k+1)\right)^2\)
and you want to use that to show that
\(\displaystyle \sum_{i=1}^{k+1} i^3= \left(\frac{1}{2}k(k+1)\right)^2\).
The reason so many "induction problems" involve sums is because
the sum to k+ 1 is just the sum to k plus one more term. Here that is
\(\displaystyle \sum_{i=1}^{k} i^3+ (k+1)^3= \left(\frac{1}{2}k(k+1)\right)^2+ (k+1)^3\).