# Error in finding sum of $1\cdot 2+3\cdot 4+ \cdots \text{to}\space n\space \text{terms}$

To find sum of the series $1\cdot 2+3\cdot 4+ \cdots \text{to}\space n\space \text{terms}$
My approach,
Let S=$1^2+2^2+3^2 + \cdots +n^2$
If $n$ is even
S=$(1-2)^2+(3-4)^2+ \cdots +[(n-1)-n]^2+2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \dfrac{n}{2}\space \text{terms})$
=$\frac{n}2+2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \dfrac{n}{2}\space \text{terms})$
And we know,
$$S=\frac{n(n+1)(2n+1)}6$$
Therefore,$$2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \frac{n}2\space \text{terms}) =\frac{n(n+1)(2n+1)}6-\frac{n}2$$
or $$2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \frac{n}2\space \text{terms})=\frac{n}2.\frac{2n^2+3n-2}3$$
or $$(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \frac{n}2\space \text{terms})=\frac{n(n+2)(2n-1)}{24}$$
But, this is the sum of $\dfrac{n}{2}$ terms. To get sum of $n$ terms, we replace $n$ by $2n$
Thus,
$$(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space n\space \text{terms})=\frac{n(n+1)(4n-1)}6$$
So, if input of $n$ is even result should be correct,
if $n=2$, sum (from formula) = $7$
However, actual sum is $14$ (=$1 \cdot 2+3 \cdot 4$).
What is the error in the above approach?
EDIT: I rechecked my calculations and also have written the steps but still the answer isn't coming right. Please tell in which step the problem lies.
Thanks for your time and patience.

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You have several errors. If you want to find the sum for even $n$, then substitute $n=2k$ and then calculate more accurately. – njguliyev Oct 26 '13 at 19:04
You are making a mistake in the 'combining the above results' part. Your sum of $\frac{n}{2}$ terms is equal to $\frac{1}{2} (S - \frac{n}{2})$, and this is not equal to $\frac{n(n+2)(2n-1)}{8}$. – Arthur Oct 26 '13 at 19:11
@njguliyev I have correctly assummed $n$ is even because only then can I pair L.H.S. – Ishan Singh Oct 27 '13 at 9:53
Your first error is "... to $n/2$ terms". That is why I suggest you to start with the substitution $n = 2k$. – njguliyev Oct 27 '13 at 10:05

Note that $$\frac{n(n + 1)(2n + 1)}{6} - \frac{n}{2} = \frac{n}{2}\frac{2n^{2} + 3n - 2}{3}$$ and if you divide this expression by $2$ you are supposed to get $$\frac{n(n + 2)(2n - 1)}{12}$$ but you get mistakenly $$\frac{n(n + 2)(2n - 1)}{24}$$ This is the source of error. If you fix this problem you will get the right formula $$(1\cdot 2) + (2\cdot 4) + \cdots + \{(2n - 1)\cdot (2n)\} = \frac{n(n + 1)(4n - 1)}{3}$$

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I am still surprised how you missed such a simple calculation mistake even after a recheck. The error should also have been obvious because you were getting answers which were half the correct value ($7$ instead of $14$). But your approach is right. – Paramanand Singh Nov 1 '13 at 14:48

$$\sum_{m=1}^n (2m-1)(2m) = \frac{1}{3}n(n+1)(4n-1)$$

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But MathGod's question is what is their error, not what is the actual formula. I would give 1/2 a vote if I could. – Casteels Oct 26 '13 at 19:06

Interesting approach and it will work for all values of n if you do it right.

The sum of first n terms in the sequence 1.2 + 3.4 + ... is calculated by the taking first sum of 2n terms of the sum of squares sequence (S in your case), subtracting n from it and taking the half.

say n = 2 sum of first 4 squares is 30 subtract 2 and take half 14.

for n = 3 sum of first 6 squares is 91 subtract 2 and take half 44.

and so on.

Your method works, however when you write "Combining the above results, we get," there you have done a calculation mistake.

There are simpler ways of doing the problem, however I like your approach.

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