How Do You Calculate the Sum of Squares from \(n+1\) to \(2n\)?

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To calculate the sum of squares from \(n+1\) to \(2n\), first correct the given formula to \(\sum_{r=1}^n r^2 = \frac{1}{6} n(n+1)(2n+1)\). Use this formula to find \(\sum_{r=1}^{2n} r^2\) and \(\sum_{r=1}^{n} r^2\). The difference between these two sums allows for the calculation of \(\sum_{r=n+1}^{2n} r^2\). Finally, multiply the result by 4 to obtain \(\sum_{r=n+1}^{2n} 4r^2\).
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I've got a very tricky series question on hand,

I'm given \sum_{r=1}^{\infty} r^2 = \frac{1}{6} n(n+1)(2n+1) and I'm asked to find \sum_{r=n+1}^{2n} 4r^2.

The change of limits and all is killing me ! any ideas on how to start? Thanks.
 
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There's an obvious typo in the "given" formula- there is no "n" on the left hand side. It should be
\sum_{r=1}^n r^2 = \frac{1}{6} n(n+1)(2n+1)

First, obvious point: ignore the "4" until you have done just \sum_{r=n+1}^{2n} r^2, then multiply by 4.

Can you use the "given" formula to find \sum_{r=1}^{2n} r^2?
Can you use the "given" formula to find \sum_{r=1}^{n} r^2?

What is the difference between them?
 
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OH yes ! thanks a lot for the help, and esp. pointing out that typo. Am able to solve now.

: )
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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