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Show $a+(a+d)+(a+2d)+\cdots+(a+nd)=a(n+1)+d\frac{n(n+1)}{2}$, where $a$ and $d$ are real numbers and $n$ is an integer.

Attempt:

I first added twice

$$a+(a+d)+(a+2d)+\cdots+(a+nd)$$

to itself in the following way:

$$[a+(a+d)+(a+2d)+\cdots+(a+nd)] + [(a+nd)+(a+(n-1)d)+(a+(n-2)d)+\cdots+a]$$

which equals $(2a+nd)+(2a+nd)+\cdots+(2a+nd)$ which are n quantities.

These $n$ quantities can then be combined to

$$2[a+(a+d)+(a+2d)+\cdots+(a+nd)]=n(2a+nd)$$

$[a+(a+d)+(a+2d)+\cdots+(a+nd)]=\frac{n(2a+nd)}{2}$ divide by $2$

After that point, I begin to screw up and don't find myself any closer to

$$a(n+1)+d\frac{n(n+1)}{2}$$

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\begin{align} S &= a+(a+d)+(a+2d)+\cdots+(a+nd)\\ &= (a+a+a+\cdots a) + (0+d+2d+3d+\cdots+ nd)\\ &= (n+1)a + d(1+2+3+\cdots+ n)\\ &= (n+1)a + d\dfrac{n(n+1)}{2} \end{align}

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Given $$ S = a+(a+d)+(a+2d)+\cdots+(a+nd) \tag 1$$

writting in reverse order,we get

$$ S = (a+nd)+(a+(n-1)d)+\cdots+a \tag 2$$

So adding these two, we get

$$2S = \left[(2a+nd)+(2a+nd)+\cdots+(2a+nd)\right] = (n+1)\cdot (2a+nd)$$

So we get $$S = \frac{(n+1)}{2}\left[2a+nd\right]$$

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  • $\begingroup$ ...product $n+1$ instead of $n$ $\endgroup$ – chenyuandong Aug 17 '15 at 3:01
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$$S:\quad (1)\quad notice\quad a+(a+nd)\quad =\quad a+d+(a+(n-1)d)\\ \quad \quad (2)\quad so\quad ,\quad the\quad number\quad of\quad limits\quad is\quad n+1\\ \quad \quad (3)\quad so\quad ,\quad a+(a+d)+(a+2d)+⋯+(a+nd)\quad =\quad (2a+nd)(\frac { n+1 }{ 2 } )\quad \\ \quad \quad \quad we\quad can\quad write\quad as\quad \rightarrow \quad a(n+1)+d\frac{n(n+1) }{ 2 } $$

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