# Set of values of n for which >10000

• December 6th 2011, 05:06 AM
Punch
Set of values of n for which >10000
Given that $\sum^n_{r=1}r(2r-1)=\frac{1}{6}n(n+1)(4n-1)$

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

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

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

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

How do I solve for n?
• December 6th 2011, 05:56 AM
CaptainBlack
Re: Set of values of n for which >10000
Quote:

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

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

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

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

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

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

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

CB
• December 6th 2011, 05:59 AM
Punch
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
$\frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}=1000$.

CB

Does that mean that I have to use a guess and check method?
• December 6th 2011, 06:04 AM
CaptainBlack
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.
• December 6th 2011, 05:50 PM
Punch
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
• December 6th 2011, 07:06 PM
CaptainBlack
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:

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

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

CB