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  • Prove the following by using the principle of mathematical induction for all n ∈ N : (1) 1+3^2+3^3+...+3^n-1=(3^n - 1)/2.)

Prove the following by using the principle of mathematical induction for all  n\in N:

Q: 1        1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}

Answers (1)

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Let the given statement be p(n) i.e.
p(n):1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}
For n = 1  we have
p(1): 1=\frac{(3^1-1)}{2}=\frac{3-1}{2}= \frac{2}{2}=1    ,   which is true

For  n = k  we have
p(k):1+3+3^2+...+3^{k-1}=\frac{(3^k-1)}{2} \ \ \ \ \ \ \ -(i)   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
p(k+1):1+3+3^2+...+3^{k+1-1}= 1+3+3^2+...+3^{k-1}+3^{k}
                                                                           = (1+3+3^2+...+3^{k-1})+3^{k}
                                                                           = \frac{(3^k-1)}{2}+3^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Using \ (i))
                                                                           = \frac{3^k-1+2.3^k}{2}
                                                                           = \frac{3^k(1+2)-1}{2}
                                                                           = \frac{3.3^k-1}{2}
                                                                           = \frac{3^{k+1}-1}{2}
Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is true for all natural numbers n

Posted by

Gautam harsolia

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