Sum Dependent on Variable: nx^2+2^2/(x+1)+3^2/(x+2)......+(n+1)^2/(x+n)=nx + n(n+3)/2

[Surreal]

For which x€R and x>0 the following equality is satisfied:
nx^2+2^2/(x+1)+3^2/(x+2)......+(n+1)^2/(x+n)=nx + n(n+3)/2
?
I found out that for x=1 the equality holds true but I'm not sure how to prove if there are any other answers.
Thanks in advance!

 

[Denis]

nx^2+2^2/(x+1)+3^2/(x+2)......+(n+1)^2/(x+n)=nx + n(n+3)/2

Where d'heck didya get that weirdo

Seems to be "sum of first n integers: n(n+1)/2" in disguise...
x = 1:
1^2/(x+0) = 1
2^2/(x+1) = 2
3^2/(x+2) = 3
...and so on

 

[Dr.Peterson]

For which x€R and x>0 the following equality is satisfied:
nx^2+2^2/(x+1)+3^2/(x+2)......+(n+1)^2/(x+n)=nx + n(n+3)/2
?
I found out that for x=1 the equality holds true but I'm not sure how to prove if there are any other answers.
Thanks in advance!

Nothing is said about the role of n; presumably it means that the equation must be satisfied for a given value of x, for all positive integers n.

So whatever x turns out to be, the equation must be true for n=1, right? Set n to 1, and solve for x; there will be two solutions. Then see if the equation holds for all n for each x.

Note something interesting: the left side is not defined for any negative integer x (since one term will require division by zero), so even if it were not stated that x must be positive, we would have a restriction.

 

[JeffM]

For which x€R and x>0 the following equality is satisfied:
nx^2+2^2/(x+1)+3^2/(x+2)......+(n+1)^2/(x+n)=nx + n(n+3)/2
?
I found out that for x=1 the equality holds true but I'm not sure how to prove if there are any other answers.
Thanks in advance!

I do not know whether the OP stll has any interest in this thread. I have been traveling and have not had time to look for a proof. I suspect the OP is correct that 1 is the only positive answer.

First, for any positive integer n, x = 1 is a solution.

\(\displaystyle \displaystyle x = 1 \implies nx^2 + \sum_{j=2}^{n + 1} \dfrac{j^2}{x + j - 1} = n * 1^2 + \sum_{j=2}^{n+1} \dfrac{j^2}{1 + j - 1} = n * 1 + \sum_{j=2}^{n+1} \dfrac{j^2}{j} =\)

\(\displaystyle \displaystyle nx + \sum_{j=2}^{n+1}j = nx + \left ( \sum_{j=1}^{n+1} j \right ) - \sum_{j=1}^{n +1} j = nx + \dfrac{(n + 1)(n + 2)}{2} - 1 = nx + \dfrac{n^2 + 3n + 2}{2} - \dfrac{2}{2}=\)

\(\displaystyle nx + \dfrac{n(n + 3)}{2}.\)

At this point I am not quite sure what exactly the problem is. But a starting point for a proof by induction is to look at the number of positive real solutions for n = 1 and n = 2. (I want look at both in case we need to deal with different analyses for odd and even numbers.)

\(\displaystyle \displaystyle n = 1, x > 0, \text { and } nx^2 + \sum_{j=2}^{n + 1} \dfrac{j^2}{x + j - 1} = nx + \dfrac{n(n + 3)}{2} \implies\)

\(\displaystyle x^2 + \dfrac{4}{x + 1} = x + 2 \implies x^3 + x^2 + 4 = x^2 + x + 2x + 2 \implies\)

\(\displaystyle x^3 - 3x + 2 = 0 \implies (x^2 - 2x + 1)(x + 2) = 0 \implies (x - 1)^2(x + 2) = 0 \implies\)

\(\displaystyle x = 1 \text { or } x = -\ 2 \implies x = 1 \ \because x > 0.\)

Time to start driving. I shall report on n = 2 later.

 

[JeffM]

For which x€R and x>0 the following equality is satisfied:
nx^2+2^2/(x+1)+3^2/(x+2)......+(n+1)^2/(x+n)=nx + n(n+3)/2
?
I found out that for x=1 the equality holds true but I'm not sure how to prove if there are any other answers.
Thanks in advance!

Well, despite considerable effort, I have not found an algebraic proof. I do not assert that an algebraic proof is impossible; indeed I am relatively confident that one exists. I merely am not clever enough to find it. Finding an algebraic proof might make a good challenge problem. (Denis likes puzzles.)

A proof using differential calculus, however, is easy.

\(\displaystyle \text {PROVE: } x = 1 \text { if } x \in \mathbb R,\ x > 0,\ f_n(x) = 0,\ n \in \mathbb Z,\ n \ge 1,\)

\(\displaystyle \displaystyle \text{and } f_n(x) = nx^2 - nx - \dfrac{n(n + 3)}{2} + \sum_{j=1}^n \dfrac{(j + 1)^2}{j + x}.\)

Proof below ignores the possibility of complex roots because x is real by hypothesis.

\(\displaystyle \displaystyle f_n(x) = nx^2 - nx - \dfrac{n(n + 3)}{2} + \sum_{j=1}^n \dfrac{(j + 1)^2}{j + x} \implies\)

\(\displaystyle \displaystyle f_n(x) = nx^2 - nx - \dfrac{n(n + 3)}{2} + \left ( \sum_{j=1}^n (j + 1)^2(j + x)^{-1} \right ).\)

\(\displaystyle \displaystyle \therefore f_n'(x) = 2nx - n + \left ( \sum_{j=1}^n (-\ 1)(j + 1)^2 (j + x)^{-2} \right ).\)

\(\displaystyle \displaystyle \therefore f_n''(x) = 2n + \left ( \sum_{j=1}^n (-\ 1)(-\ 2)(j+ 1)^2 (j + x)^{-3} \right ).\)

\(\displaystyle \displaystyle \text {Or } f_n'(x) = 2nx - n - \sum_{j=1}^n \dfrac{(j + 1)^2}{(j + x)^2} \text { and } f_n''(x) = 2n + 2 * \sum_{j=1}^n \dfrac{(j + 1)^2}{(j + x)^3}.\)

\(\displaystyle \therefore x > 0 \implies f_n''(x) > 0.\)

\(\displaystyle \displaystyle \text {And } f_n'(1) = 2n(1) - n - \sum_{j= 1}^n \dfrac{(j + 1)^2}{(j + 1)^2} = n - \sum_{j=1}^n 1 = n - n = 0.\)

\(\displaystyle \therefore x > 0 \text { and } x \ne 1 \implies f_n(x) > f_n(1).\)

\(\displaystyle \displaystyle f_n(1) = n * 1^2 - n * 1 - \dfrac{n(n + 3)}{2} + \sum_{j= 1}^n \dfrac{(j + 1)^2}{j + 1} = \left ( \sum_{j=1}^n j + 1 \right ) - \dfrac{n^2 + 3n}{2} \implies \)

\(\displaystyle \displaystyle f_n(1) = \left ( \sum_{j=1}^n j \right ) - \dfrac{n(n + 3)}{2} + \left ( \sum_{j= 1}^n 1 \right ) \implies\)

\(\displaystyle f_n(1) = \dfrac{n(n + 1)}{2} - \dfrac{n(n + 3)}{2} + n = \dfrac{n^2 + n - n^2 - 3n + 2n}{2} = \dfrac{0}{2} = 0.\)

\(\displaystyle \therefore x > 0 \text { and } x \ne 1 \implies f_n(x) > 0.\)

\(\displaystyle \therefore x > 0 \text { and } x \ne 1 \implies f_n(x) \ne 0.\)

\(\displaystyle \text { not } f_n(x) \ne 0 \implies \text { not } \{ x > 0 \text { and } x \ne 1 \}.\)

\(\displaystyle \therefore f_n(x) = 0 \implies x \not > 0 \text { or } x = 1.\)

\(\displaystyle \text {But, by hypothesis, } x > 0.\)

\(\displaystyle \therefore x = 1. \text { Q.E.D.}\)