Are you smart in Summation Problems?

Algebra Level 2

$\large\displaystyle\sum_{i=1}^{n} \dfrac{x+i}{3i+2}=\displaystyle\sum_{i=1}^{n} \dfrac{i+1}{x+3i+1}$

Solve for real ${x}$ .

The answer is 1.

2 solutions

Chew-Seong Cheong
Jul 11, 2015

$\begin{aligned} \sum_{i=1}^n \frac{x+i}{3i+2} & = \sum_{i=1}^n \frac{i+1}{x+3i+1} \end{aligned}$

For the equation to be true for all $i$ and $n$ :

$\begin{aligned} \frac{x+i}{3i+2} & = \frac{i+1}{x+3i+1} \\ (x+i)(x+3i+1) & = (i+1)(3i+2) \\ x^2 + (i + 3i+1)x 3i^2+ i & = 3i^2 +2i+ 3i + 2 \\ x^2 + (4i+1)x -(4i+2) & = 0 \\ (x-1)(x+4i+2) & = 0 \end{aligned}$

Since $x+4i + 2 \ne 0$ for all $i$ , then the only solution is $x = \boxed{1}$ . We see that when $x = 1$ , $LHS \equiv RHS$ .

Moderator note:

So yes, you showed that in the particular case of $i = 1, n = 1$ , we must have $x = 1$ .

What about all other cases? Is $x = 1$ a solution? Is $x = 1$ the only real solution?

Is there a solution for the other cases? I would like to know. Thanks

Leonardo Martins Bianco - 5 years, 11 months ago

There won't be any other cases. If you do some algebraic manipulation the equation will become as $(x-1)\times \sum _{ i=1 }^{ n }{ \frac { x+4i+2 }{ (3i+2)(x+3i+1) } =0 } \\ if\quad x\quad \neq \quad 1\quad then\\ \sum _{ i=1 }^{ n }{ \frac { x+4i+2 }{ (3i+2)(x+3i+1) } =0 } \\ for\quad large\quad n\quad \sum _{ i=1 }^{ n }{ \frac { x+4i+2 }{ (3i+2)(x+3i+1) } >0 }$ Hence the only solution to this problem is $x=1$

Som Ghosh - 5 years, 11 months ago

For large $n$ ? What happens when $n$ is not large, and $x$ is a negative number?

Satyajit Mohanty - 5 years, 11 months ago

@Satyajit Mohanty – There is a possibility of getting a negative number, hence possibility of getting other solutions too.

Som Ghosh - 5 years, 11 months ago

@Som Ghosh – Yeah! But you need to prove the impossibility of your so called possibility :D

Satyajit Mohanty - 5 years, 11 months ago

Thank you. Really nice solution!

Leonardo Martins Bianco - 5 years, 10 months ago

I have changed the last paragraph to explain that the answer is unique.

Chew-Seong Cheong - 5 years, 11 months ago

Department 8
Jul 14, 2015

I am seriously sorry for this.By comparing both sides we will see the minimum value of $x$ is:- $x+i=i+1\\ and\quad \\ x+3i+1=3i+2$

From this we see $\boxed{x=1}$

You really have an Indian Brain then :D

Satyajit Mohanty - 5 years, 11 months ago

bhai tukka toh sahi mara h maine bhi aise hi kia hai but i know imaginary solution doesn't exist.

Shashank Rustagi - 5 years, 11 months ago

Are you smart in Summation Problems?

The answer is 1.

2 solutions

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