# Lim n tends to infinty [{(1^2+2^2+3^2+...n^2)(1^3+2^3+3^3+...n^3)}/(1^6+2^6+3^6+...n^6)]=??

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#### Explanation

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#### Explanation:

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Dec 13, 2017

$= \frac{7}{12}$

#### Explanation:

We know that 1^r+2^r+3^r+…+n^r can be expressed as a formula in the form of a polynomial with degree $r + 1$.

So, 1^2+2^2+3^2+…+n^2=an^3+bn^2+cn+d for all $n$ with certain $a , b , c , d$. This can be solved by setting $n = 0 , 1 , 2 , 3$ to get a system of linear equations:
$\left\{\begin{matrix}d = 0 \\ a + b + c + d = 1 \\ 8 a + 4 b + 2 c + d = 5 \\ 27 a + 9 b + 3 c + d = 14\end{matrix}\right.$

Solving this, we get $a = \frac{1}{3} , b = \frac{1}{2} , c = \frac{1}{6} , d = 0$.

Thus, 1^2+2^2+3^2+…+n^2=1/3n^3+1/2n^2+1/6n=(2n^3+3n^2+n)/6.

Use a similar process to find that 1^3+2^3+3^3+…+n^3=(n^4+2n^3+n^2)/4 and 1^6+2^6+3^6+…+n^6=(6n^7+21n^6+21n^5-7n^3+n)/42.

Then, the original problem becomes
${\lim}_{n \to \infty} \frac{\left(\frac{2 {n}^{3} + 3 {n}^{2} + n}{6}\right) \left(\frac{{n}^{4} + 2 {n}^{3} + {n}^{2}}{4}\right)}{\frac{6 {n}^{7} + 21 {n}^{6} + 21 {n}^{5} - 7 {n}^{3} + n}{42}}$
$= {\lim}_{n \to \infty} \frac{42 \left(2 {n}^{3} + 3 {n}^{2} + n\right) \left({n}^{4} + 2 {n}^{3} + {n}^{2}\right)}{24 \left(6 {n}^{7} + 21 {n}^{6} + 21 {n}^{5} - 7 {n}^{3} + n\right)}$
$= \frac{7}{12}$.

See the other solution I posted for a method involving integration but is considerably easier.

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Dec 13, 2017

$= \frac{7}{12}$

#### Explanation:

((1^2+2^2+3^2+...+n^2)(1^3+2^3+3^3+….+n^3))/(1^6+2^6+3^6+…+n^6)
=(n^5((1/n)^2+(2/n)^2+(3/n)^2+...+1^2)((1/n)^3+(2/n)^3+(3/n)^3+…+1^3))/(n^6((1/n)^6+(2/n)^6+(3/n)^6+…+1^6))

$= \frac{\frac{1}{n} {\sum}_{i = 0}^{n} {\left(\frac{i}{n}\right)}^{2} \frac{1}{n} {\sum}_{i = 0}^{n} {\left(\frac{i}{n}\right)}^{3}}{\frac{1}{n} {\sum}_{i = 0}^{n} {\left(\frac{i}{n}\right)}^{6}}$

As $n \to \infty$, using Riemann sums, we find
$= \frac{{\int}_{0}^{1} {x}^{2} \setminus \mathrm{dx} \setminus {\int}_{0}^{1} {x}^{3} \setminus \mathrm{dx}}{{\int}_{0}^{1} {x}^{6} \setminus \mathrm{dx}}$
$= \frac{{\left[{x}^{3} / 3\right]}_{0}^{1} {\left[{x}^{4} / 4\right]}_{0}^{1}}{{x}^{7} / 7} _ {0}^{1}$
$= \frac{\frac{1}{3} \cdot \frac{1}{4}}{\frac{1}{7}}$
$= \frac{7}{12}$

See the other solution I posted for an approach that does not resort to integration.

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