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My question actually come from this:

Intuitively understanding $\sum_{i=1}^ni={n+1\choose2}$

I was once examining the sum of an A.P series with first term $a$ and common difference $d$ . And this question mentioned above reminded me of same.

$S_n=\frac{n}{2}(2a+(n-1)d)=a n+\frac{n(n-1)}{2}d=a \cdot\;^{n}C_1+\;^{n}C_{2} \cdot d$

Is there any special significance or intuition behind this result? Why does those combinations appear? can it be extended? Or it's pure coincidence?

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Someone
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  • Hint. You have separated the sums of the $a$ terms and the $d$ terms. Now think about $\sum_1^n (a + bi + ci^2)$. That might give you a generalization - maybe not the one you expect. – Ethan Bolker May 08 '15 at 16:38
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    Ah I know, it's quite similar to the case taking $a$ and $d$ common but i have wanted to know is there any other way of seeing it, maybe in terms of choosing $a$'s and $d$'s. – Someone May 08 '15 at 16:41

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The expression for an A.P. follows easily from the expression for the sum of numbers from $1$ to $n$:

$$S_n \equiv \sum_{k=1}^{n} \left( a+(k-1)d \right)$$ We can break up the sum, that is, take out the $a$ part and treat it separately.

$$S_n =\sum_{k=1}^{n} \left( a \right) + \sum_{k=1}^{n} \left( (k-1)d \right)$$

The first term is easy; you add $n$ copies of a constant to get $na$. We can pull the constant $d$ outside the second sum:

$$S_n =na + d\sum_{k=1}^{n} \left( (k-1) \right) = na + d\sum_{j=0}^{n-1} \left( j \right)$$

Finally, we apply our sum of integers formula $\sum_{j=0}^m j = \sum_{j=1}^m j = \binom{m+1}{2}$ to get

$$\sum_{j=0}^{n-1} \left( j \right) = \binom{(n-1)+1}{2} $$ So $$ S_n = an + d\binom{n}{2} = a\binom{n}{1} + d\binom{n}{2} $$

This can be extended for sequences where the first difference is itself an arithmetic progression, for example, $$2+3+6+ 11+ 18 + 27 +\cdots $$ in which case there will now be an additional term in the answer involving $\binom{n}{3} $.

Mark Fischler
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