Simple induction proof (1 Viewer)

Run hard@thehsc · Mar 6, 2022

the proof is ease, but I am messing up somewhere in the last step regarding the factorisation - thanks for the help:

5uckerberg · Mar 6, 2022

So you're on the n=k+1 step. Alright.

Start with this
$\sum_{r=1}^{k}\left(r+1\right)\times{2^{r}}=n\times{2^{n+1}}$

For the n=k+1 case
$\sum_{r=1}^{k+1}\left(r+1\right)\times{2^{r}}=n\times{2^{n+1}}$

Notice that for the n=k case
$\sum_{r=1}^{k}\left(r+1\right)\times{2^{r}}=k\times{2^{k+1}}$ .

Thus, for the n=k+1 case
$\sum_{r=1}^{k+1}\left(r+1\right)\times{2^{r}}=\left(k+1\right)\times{2^{k+2}}=k\cdot{2^{k+2}}+2^{k+2}$

Working with the LHS
$\sum_{r=1}^{k+1}\left(r+1\right)\times{2^{r}}$ is just
$k\times{2^{k+1}}+\left(k+2\right)\times{2^{k+1}}=k\cdot{2^{k+2}}+2^{k+2}$ using the assumption made from the n=k statement

Cleaning this up on the LHS we will have
$k\cdot{2^{k+1}}+k\cdot{2^{k+1}}+2^{k+2}$
That simply gives us
$k\cdot{2^{k+2}}+2^{k+2}$
equalling the RHS.

The rest is up to you. But I recommend to write your induction proof as if you are completing an essay according to Bill Pender.

yanujw · Mar 6, 2022

Assuming you could do the 'prove true for n=1' part...

Assume true for n=k;
2^r(r+1) + 2^(r+1)(r+2) + ... + 2^k(k+1) = k2^(k+1)

Prove true for n=k+1;
LHS = 2^r(r+1) + 2^(r+1)(r+2) + ... + 2^k(k+1) + 2^(k+1)(k+2)
= k2^(k+1) +2^(k+1)(k+2) by induction hypothesis
= (2k+2)2^(k+1)
=(k+1)2^(k+2)
= RHS when n=k+1

True for n=k+1 when true for n=k

By math induction, statement is true for postive integers n.

Simple induction proof (1 Viewer)

Run hard@thehsc

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5uckerberg

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yanujw

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