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Algebra Level 4

$Given :: \ \displaystyle S_{n}=\sum_{r=1}^n t_{r}= \frac{n(2n^{2}+9n+13)}{6}$ $Find \quad \displaystyle \sum_{r=1}^\infty \frac{1}{r.\sqrt{t_{r}}}$

1 solution

Arif Ahmed
Oct 3, 2014

The Right Hand Side term is cubic so the ${t}_{r}$ must be of degree $2$

So one could begin by thinking over known forms such as :

$\sum_{r=1}^{n}{{r}^{2}} \quad = \quad \frac {n(n+1)(2n+1)}{6}$

$\sum_{r=1}^{n}{r} \quad = \quad \frac {n(n+1)}{2}$

$\sum_{r=1}^{n}{1} \quad = \quad n$

So , ${t}_{r} = A{r}^{2} + Br + C$

Therefore , using the known forms and comparing the result with Right Hand Side cubic expression one should get the following system of equations :

$2A = 2$

$3A + 3B = 9$

$A + 3B + 6C = 13$

Solving one will get $A = 1 , B = 2 , C = 1$

Therefore , ${t}_{r} = {(r+1)}^{2}$

Second part is easy : $\sum_{r=1}^{\infty}{\frac {1}{r(r+1)}}$ $= \sum_{r=1}^{\infty}{\frac{1}{r} - \frac{1}{r+1}}$

This telescopes to $1$ .

Another way to get $t_{r}$ is to see that
$t_{r}=S_{r+1}-S_{r}$
and the rest follows....

Ishan Singh - 6 years, 8 months ago

I too did it your way , just completing it

$t_r = S_r - S_{r -1}$

$t_r = \dfrac{r(2r^2 + 9r + 13)}{6} - \dfrac{(r-1)(2(r-1)^2 + 9(r-1) + 13)}{6}$

$t_r = \dfrac{2(r^3 - (r - 1)^3) + 9(r^2 - (r - 1)^2) + 13(r - r + 1)}{6}$

$t_r = \dfrac{ 6r^2 + 12r + 6}{6} = (r +1)^2$

$\dfrac{1}{r\sqrt{t_r}} = \dfrac{1}{r(r + 1)} = \dfrac{1}{r} - \dfrac{1}{r +1}$

U Z - 6 years, 4 months ago

Ishaan nice one.

Nivedit Jain - 4 years, 6 months ago