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Thread: Series

  1. #1
    Member srirahulan's Avatar
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    Prove That $\displaystyle |U_{n}-U_{n-1}|=\frac{1}{3^{n-1}}$, for all$\displaystyle n\epsilon Z^{+}$,Where$\displaystyle U_{1}=1,U_{2}=2 $and$\displaystyle U_{n}=\frac{1}{3}[U_{n-2}+2U_{n-1}]$ When n=3,4,5............. by using the method of Principle of Mathematical Induction. This problem, I firstly show that this case is true when n=1 and after that consider that when n=p $\displaystyle (p\epsilon Z^{+})$ the case was true but after that how can i insert (p+1) and why they give$\displaystyle U_{n}$...please help me
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    Re: Series

    Quote Originally Posted by srirahulan View Post
    Prove That $\displaystyle |U_{n}-U_{n-1}|=\frac{1}{3^{n-1}}$, for all$\displaystyle n\epsilon Z^{+}$,Where$\displaystyle U_{1}=1,U_{2}=2 $and$\displaystyle U_{n}=\frac{1}{3}[U_{n-2}+2U_{n-1}]$ When n=3,4,5............. by using the method of Principle of Mathematical Induction. This problem, I firstly show that this case is true when n=1 and after that consider that when n=p $\displaystyle (p\epsilon Z^{+})$ the case was true but after that how can i insert (p+1) and why they give$\displaystyle U_{n}$...please help me
    Actually the base case is $\displaystyle n=3$.
    Assume $\displaystyle K>3$ and $\displaystyle |U_{K}-U_{K-1}|=\frac{1}{3^{K-1}}$ is true.

    Then show that $\displaystyle |U_{K+1}-U_{K}|=\frac{1}{3^{K}}$ is also true.
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