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# proof by induction

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so far i tried subbing in r for Un but i have no idea how to do it

Jan 10, 2019

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Solution for part b:

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Jan 10, 2019
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thank you for the detailed explanation

can you explain why you did Ur + U n+1 at the first part of ii please its the only part i didnt catch on to

YEEEEEET  Jan 10, 2019
edited by YEEEEEET  Jan 10, 2019
#3
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"can you explain why you did Ur + U n+1 at the first part of ii please "

Think of the summation as follows:  $$\sum_{k=1}^nu_k=u_1+u_2+...+u_n$$

Hence $$\sum_{k=1}^{n+1}u_k=u_1+u_2+...+u_n+u_{n+1}$$  which is $$\sum_{k=1}^{n+1}u_k=\sum_{k=1}^nu_k+u_{n+1}$$

Hope that makes it clear.

Alan  Jan 10, 2019
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perfect! i forgot about that haha thank you

YEEEEEET  Jan 10, 2019