Solve this problem that involves induction

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Homework Statement
See attached.
Relevant Equations
Mathematical induction
Hello,
This is the attachment, the steps to solution are pretty clear. I guess there is a mistake on the highlighted part that prompts this thread.

1756810436257.webp




Ought to be ##3^{n+1} (n+2)-6## and not ##3^n(n+2)-6##. Unless i missed something, on another note, i find the first method (induction) better than second one (method of differences).
 

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Another method is to define [tex]f(x) = \sum_{k=1}^n x^k = \frac{x^{n+1}-x}{x-1}[/tex] so that [tex]xf'(x) = \sum_{k=1}^n kx^k[/tex] and we are looking to calculate [tex] \sum_{k=1}^n 3^k(2k+5) = 6f'(3) + 5f(3).[/tex]
 
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chwala said:
I guess there is a mistake on the highlighted part that prompts this thread.

Ought to be ##3^{n+1} (n+2)-6## and not ##3^n(n+2)-6##. Unless i missed something

I don't see any highlighted part.
 
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Mark44 said:
I don't see any highlighted part.

There is a faint red box around the [itex]3^n[/itex] in the last expression on the right. It should read [itex]3^{n+1}[/itex] as in the line above.
 
pasmith said:
There is a faint red box around the [itex]3^n[/itex] in the last expression on the right. It should read [itex]3^{n+1}[/itex] as in the line above.
I see it now after you pointed it out, but that's very faint.
 
pasmith said:
Another method is to define [tex]f(x) = \sum_{k=1}^n x^k = \frac{x^{n+1}-x}{x-1}[/tex] so that [tex]xf'(x) = \sum_{k=1}^n kx^k[/tex] and we are looking to calculate [tex] \sum_{k=1}^n 3^k(2k+5) = 6f'(3) + 5f(3).[/tex]
This is new to me, @pasmith. Can you continue with the next lines? Cheers, man.
 
chwala said:
This is new to me, @pasmith. Can you continue with the next lines? Cheers, man.
Express ## f(3) ## and ## f’(3) ## in terms of ## n ## by using $$ f(x)=\frac{x^{n+1}-x}{x-1} $$ and $$ f'(x)=\frac{d}{dx}(\frac{x^{n+1}-x}{x-1}) $$.
 
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Mark44 said:
I don't see any highlighted part.
@Mark i re-uploaded, i too could not see the highlighted part :biggrin:
 

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