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Determine formulas for given sums

Consider the formula

By differentiating, determine formulas for the following:

1. .
2. .

1. We observe that

Thus, using the given formula we have,

2. (Note: There’s an alternate solution to this in the comments that is more direct than this one. Definitely worth taking a look at that alternative.)
Taking derivatives of both sides of the equation we derived in part (a),

Then, looking at the second term in the sum on the left, we have,

Thus, plugging this back into the expression above,

3. ( Note: This is a lot of algebra. I included as many steps as I thought appropriate, but that still has a lot going on in each step. If there is something that isn’t clear please leave a comment. Also, given all of these equations it seems likely I’ve made at least one typo / mathematical error. If you find one, please point it out.)

1. Anonymous says:
*** QuickLaTeX cannot compile formula:
\begin{align*} 1+ 2x + \cdots + (n-1)x^{n-2} &= (1 + x + \cdots + x^n)' - nx^{n-1} \\ &= \left( \frac{x^{n+1}-1}{x-1} \right)' - nx^{n-1} \\ &= \frac{nx^{n+1} - (n+1)x + 1}{(x-1)^2} - nx^{n-1} \\ &= \frac{(n+1)x^n - nx^{n-1} + 1}{(x-1)^2}. \end{align*}

*** Error message:
Cannot connect to QuickLaTeX server: cURL error 60: Peer's Certificate issuer is not recognized.
Please make sure your server/PHP settings allow HTTP requests to external resources ("allow_url_fopen", etc.)
These links might help in finding solution:
http://wordpress.org/extend/plugins/core-control/
http://wordpress.org/support/topic/an-unexpected-http-error-occurred-during-the-api-request-on-wordpress-3?replies=37

Could someone explain to me the equalities between the last two expressions here? I couldn’t figure it out myself and I tried plugging them into a calculator but couldn’t make them equal.

• Anonymous says:

The two equations in part b with an (x-1)^2 in the denominator (one of them is subtracting an nx^n-1 term that’s outisde of the fraction, while in the next equation everything is divided by (x-1)^2

• Anonymous says:

I think it should be a -(n+1)x^n in the first equation, and a (n-1)x^n in the second one (instead of the (n+1)x^n we have now). You can get to the final equality (for 1 +2x+…+(n-1)x^n-2) either by plugging in (n-1) for n for the solution in part a, or by doing some algebra on the equation Rori found (using (n+1)x^n instead of (n+1)x, since that’s what we have in the original equality)

2. Ganishk says:

Ans. to b)

Let the result of a) be $S_1$ and we need to find $S_2$

Since each term in b) is obtained from a) by multiplying x with the derivative of $xS_1$ .

Hence $S_2 = x D_x $$xS_1$$ \\ = x^2 D_xS_1 + xS_1$
Now by finding the derivative of $S_1$ wrt x we get,

$D_x $$S_1$$ = \dfrac{ $$n+1$$^{2_{\_}}x^{n-1}}{x-1} – \dfrac{2}{x-1}S_1$

Where $n^{2_{_2}}$ denotes the Pochammer symbol from quantum Calculus for falling factorial

Now solving for $S_2$ we get,
$S_2 = \dfrac{$$n^2+n$$x^{n+1}}{x-1} -\dfrac{x^2+x}{x-1}S_1$

I hope this method is useful and short as this uses the result of a) to simplify the derivative and please forgive me for my latex errors as I was practicing latex for these last 2 months of Covid -19 quarantine.

• Ganishk says:

Sorry I can’t edit that comment in which I forgot put the latex page command

Ans. to b)

Let the result of a) be and we need to find

Since each term in b) is obtained from a) by multiplying x with the derivative of .

Hence

Now by finding the derivative of wrt x we get,

Where denotes the Pochammer symbol from quantum Calculus for falling factorial

Now solving for we get,

I hope this method is useful and please forgive me for my latex errors as I was practicing latex for these last 2 months of Covid -19 quarantine

• Ganishk says:

Ans. to b)

Let the result of a) be and we need to find

Since each term in b) is obtained from a) by multiplying x with the derivative of .

Hence

Now by finding the derivative of wrt we get,

Where denotes the Pochammer symbol from quantum Calculus for falling factorial

Now solving for we get,

I hope this method is useful

3. tasos says:

For the part b,

This last step uses the result of part (a).

This provides a direct solution. Hopefully there’s no typo.

• RoRi says:

Hi, thanks! I retyped your comment into latex so that it’s easier to read. Let me know if I made a mistake. This is a nice solution, that I think works also. I’ll leave the solution I have up there and put a note to look at the comments for this alternative. (I’m not anxious to actually take that derivative since I’ll definitely make an algebra mistake.)