The sum of $n$ terms in $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots$ I am just confused, considering we can take $1 \cdot 2$ as the first term then we get the $n$th term as $n(n+1)$ so the sum of $n$ terms would be $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(n+1)}{2}$ but let's assume $0 \cdot 1$ as the first term then the $n$th term becomes $(n-1) \cdot n$ and so it's summation becomes $\frac{n(n+1) (2n+1)}{6} - \frac{n(n+1)}{2}$ but it's the same summation as above but the results are different what am I doing wrong?
 A: The results are same.
The first $n$ terms of first series is different from first $n$ terms of second series. Thus, You'll get different answers if you plug in same $n$ . In fact, you'll have to plug $n+1$ terms in second to get same as that of $n$ terms of first.
A: This reduces the problem to the evaluation of a second derivative.
$$
   1\cdot 2 + 2\cdot 3+ 3\cdot 4+ \cdots + n\cdot (n+1)
   = \sum_{k=1}^{n} k(k+1) \\
   = \left.\frac{d^2}{dx^2}(1+x+x^2+x^3+\cdots+x^{n+1})\right|_{x=1} \\
   = \left.\frac{d^2}{dx^2}\frac{x^{n+2}-1}{x-1}\right|_{x=1}
$$
A: $S_1=1⋅2+2⋅3+3⋅4+\dots$
$T_n=n(n+1)$
$S_1=\frac{n(n+1)(2n+1)}{6}\bf+\frac{n(n+1)}{2}$
$S_2=1⋅2+2⋅3+3⋅4+\dots$
$T_n=(n-1)n$
$S_2=\frac{n(n+1)(2n+1)}{6}\bf-\frac{n(n+1)}{2}$
As the bold part indicates, the summation isn't the same.
Let's put $\bf n=1$:
$S_1=1⋅2=2$
$S_1=\frac{1(2)(3)}{6}\bf+\frac{1(2)}{2}=1+1=2$
$S_2=0⋅1=0$
$S_2=\frac{1(2)(3)}{6}\bf-\frac{1(2)}{2}=1-1=0$
Let's put $\bf n=1$:
$S_1=1⋅2+2⋅3=8$
$S_1=\frac{2(3)(5)}{6}\bf+\frac{2(3)}{2}=5+3=8$
$S_2=0⋅1+1⋅2=2$
$S_2=\frac{2(3)(5)}{6}\bf-\frac{2(3)}{2}=5-3=2$
So, what you have done is totally correct and verifies.
Whenever there is confusion like such that you have, working out for small values of $n$ which we can ourselves calculate helps to make things clearer.
A: In first case. ${\color{green}S_n}=\frac{n(n+1)(n+2)}{6}+\frac{n(n+1)}{2}={\color{red}{\frac{n(n+1)(n+2)}{3}}}$
In second case, $S'_{n+1}=\frac{(n+1)(n+2)(2n+3)}{6}-\frac{(n+1)(n+2)}{2}={\color{red}{\frac{n(n+1)(n+2)}{3}}}={\color{green}S_n}$
