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?
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} $.