#can someone help correct my method
28 messages · Page 1 of 1 (latest)
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i got a = 16 and b = 10, which are both twice the correct answer. the correct answer is a = 8 and b = 5
idk what i did wrong
ah, okay.
How did you get this guy?
ah, i see where you went wrong now.
$\frac{1}{3}n(2n+1)(2n-1)+2n(n+1)(2n+1)$
flying green people eater
the conversion from this step to the next is wrong
The sum was named wrong
the sum of the two parts is not S_n but S_2n
o
Here's why
i think brain is freid
$S_{2n} = \sum_{k=1}^{2n} (2 + (-1)^k)k^2$
Rion
From here, you reformulate this sum by grouping the terms 2r and 2r-1 together
$S_{2n} = \sum_{r=1}^{n} \left((2r-1)^2 + 3(2r)^2\right) = \left(\sum_{r=1}^{n} (2r-1)^2\right) + \left( \sum_{r=1}^{n} 3(2r)^2 \right)$
Rion
thank you
Nice