<s>[INST] 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.
 [/INST] 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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