Set of values of n for which >10000

Given that $\displaystyle \sum^n_{r=1}r(2r-1)=\frac{1}{6}n(n+1)(4n-1)$

Find the set of values of $\displaystyle n$ for which $\displaystyle \sum^n_{r=1}(6r^2-3r+2^r)>10000$

$\displaystyle \sum^n_{r=1}(6r^2-3r+2^r)=3\sum^n_{r=1}r(2r-1)+\sum^n_{r=1}2^r$

$\displaystyle =3[\frac{1}{6}n(n+1)(4n-1)]+\frac{2(1-2^n)}{1-2}$

$\displaystyle =\frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}$

How do I solve for n?

Re: Set of values of n for which >10000

Quote:

Originally Posted by

**Punch** Given that $\displaystyle \sum^n_{r=1}r(2r-1)=\frac{1}{6}n(n+1)(4n-1)$

Find the set of values of $\displaystyle n$ for which $\displaystyle \sum^n_{r=1}(6r^2-3r+2^r)>10000$

$\displaystyle \sum^n_{r=1}(6r^2-3r+2^r)=3\sum^n_{r=1}r(2r-1)+\sum^n_{r=1}2^r$

$\displaystyle =3[\frac{1}{6}n(n+1)(4n-1)]+\frac{2(1-2^n)}{1-2}$

$\displaystyle =\frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}$

How do I solve for n?

Get your calculator out and try a few values

You may need to prove that there is but a single positive root to

$\displaystyle \frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}=1000$.

CB

Re: Set of values of n for which >10000

Quote:

Originally Posted by

**CaptainBlack** Get your calculator out and try a few values

You may need to prove that there is but a single root to

$\displaystyle \frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}=1000$.

CB

Does that mean that I have to use a guess and check method?

Re: Set of values of n for which >10000

Quote:

Originally Posted by

**Punch** Does that mean that I have to use a guess and check method?

That would do it, yes. But don't forget you need to show this has a single positive root.

Re: Set of values of n for which >10000

Quote:

Originally Posted by

**CaptainBlack** That would do it, yes. But don't forget you need to show this has a single positive root.

Does that mean I have to show b^2-4ac=0? But how do i go about doing that for an equation with a power

Re: Set of values of n for which >10000

Quote:

Originally Posted by

**Punch** Does that mean I have to show b^2-4ac=0? But how do i go about doing that for an equation with a power

No, what you need to do is show that:

$\displaystyle f(x)=\frac{1}{2}x(x+1)(4x-1)-2+2^{x+1}-1000$

is monotonic and changes sign between $\displaystyle x=0$ and say $\displaystyle x=1000$, then since $\displaystyle f(x)$ is continuous it has a single root and it lies between $\displaystyle 0$ and $\displaystyle 1000$.

CB