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Given the general sequence $a_n = 4n^2 +6n - 4$, calculate the difference $(a_2 + a_4 + ... + a_{100}) - (a_1 + a_3 + ... + a_{99})$.

If we look closely the difference can be rewritten as the sum of the differences between every element:

$(a_2 - a_1) + (a_4 - a_3) + ... +(a_{100} - a_{99})$

We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$: $$c_n = a_{n+1} - a_n = [4(n+1)^2 +6(n+1) - 4] -[4n^2 +6n - 4] = 8n + 10$$ The new sequence is an arithmetic one, where $c_1 = 18$, $d = 8$, and has 50 elements. Using the summation formula for an arithmetic sequence, we get: $$C_n = \frac{c_1+c_{n}}{2}\cdot n\rightarrow C_{50} = \frac{c_1+c_{50}}{2}\cdot 50 = \frac{18 + 410}{2}\cdot 50 = \boxed{10700}$$ This appears to be wrong however, as the correct answer says 20500.

Where have I gone wrong?

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    $\begingroup$ $6(n+1)\neq 6n+1$, so your expression for $a_{n+1}$ is wrong. Ah, you just have corrected it...may be there are more. $\endgroup$ – Dietrich Burde Jan 21 at 19:12
  • $\begingroup$ @DietrichBurde Thanks, but this is just a typographical error, my notes were different. $\endgroup$ – daedsidog Jan 21 at 19:14
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    $\begingroup$ Last element is not $c_{50}$ ! $\endgroup$ – Damien Jan 21 at 19:19
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We can turn this information into a new sequence, $c_n = a_{n+1} - a_n$:

Herein lies the problem. What you ultimately end up doing is finding $c_1 + ... + c_{50}$, but notice:

$$c_1 + c_2 = (a_2 - a_1) + (a_3 - a_2) = a_3 - a_1$$

Another example:

$$c_1 + ... + c_5 = (a_2 - a_1) + (a_3 - a_2) + ... + (a_6 - a_5) = a_6 - a_1$$

Basically, the sum over $c_n$ ends up being a telescoping sum of sorts, and you can see by now

$$c_1 + ... + c_{50} = a_{51} - a_1$$

which is not the sum you wanted to compute. Indeed, plugging in the appropriate values for $n$ shows $a_{51} - a_1 = 10,700$, your erroneous result.

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  • $\begingroup$ Ahh. Thank you. $\endgroup$ – daedsidog Jan 21 at 19:24
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$$\sum_{r=1}^{50}(a_{2r}-a_{2r-1})=\sum (4(4r-1)+6)=?$$

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