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tabish
:
prove by mathematical induction for all nEN , 1+3+3^2+.......3^n-1= 3^n-1÷2 (here 3 power n and -1 )
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February 27, 2017 at 11:22am
Little Sky
:
Your interpreted question:[deqn]3^0+...+3^{n-1}=\frac{3^n-1}{2}[/deqn]We need to form a base case here. Think of the mathematical induction as forming a chain. We need to prove that the first case is true, which is the existence of the 1st member of chain is true. We need to make an assumption that the method of forming (k) member the chain work (you can think of k being the second member), and use that method to form the next member (k+1) of the chain (You can think of k+1 being the third one). Once the next member is formed, we have proved that our assumption of the method of forming the chain works, which of course the same method will work for immediate next and the following n members. Base Case: when n = 1, prove that L.H.S. = R.H.S. You can work out that both side equals to 1. Assumption Case: Assume n = k is true. [deqn]3^0+...+3^{k-1}=\frac{3^k-1}{2}[/deqn] Prove that n = k + 1 is true. i.e. prove that L.H.S. = R.H.S. by making use of the assumption. You can approach from the L.H.S., making use of the assumption case to reach your R.H.S. So, [deqn]\begin{align}L.H.S.&=3^0+...+3^{(k+1)-1}\\&=3^0+...+3^k\\&=3^0+...+3^{k-1}+3^k\\&=\frac{3^k-1}{2}+3^k\\&=\frac{3^k-1}{2}+\frac{(2)3^k}{2}\\&=\frac{3^k-1+(2)3^k}{2}\\&=\frac{3^k+(2)3^k-1}{2}\\&=\frac{3^k(1+2)-1}{2}\\&=\frac{3^k(3)-1}{2}\\&=\frac{3^{(k+1)}-1}{2}\\&=R.H.S~[\text{proven}]\end{align}[/deqn]
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February 27, 2017 at 1:22pm
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