evaluation of this sum?

Question

$$\sum\limits_{j=i}^{n} j = i + (i+1) + (i+2) + \cdots + (i+k)$$ note: above $(i+k) = n$ $$\\$$ now $$\sum\limits_{j=i}^{n} j = i \cdot(k+1) + \frac{(k+1)k}{2} = \frac{(k+1)(2i+k)}{2} $$ so my question is where does the $i \cdot (k+1)$ come from?? Also, $$\frac{(k+1)(2i+k)}{2} = \frac{(n-i+1)(2i+n-i)}{2} = \frac{(n-i+1)(n+i)}{2}$$ How does the $$\frac{(k+1)(2i+k)}{2} = \frac{(n-i+1)(2i+n-i)}{2} ??$$

Answer 1

The idea here is that each of the $k+1$ terms in your summation contain an $i$, so we can pull those out to get: $$ i + (i+1) + (i + 2) + \dots + (i + k) = (i + i + \dots + i) + (1 + 2 + \dots + k) $$ The first summation is just $i$, repeated $k+1$ times. The second summation is given by $\frac{k(k+1)}{2}$. For your second question, this is just the substitution $k = n-i$.

Alternatively, we could rewrite the sum as the following: $$ \sum_{j=i}^n j = \sum_{j = 0}^{n - i} (i + j) = \sum_{j = 0}^{n - i} i + \sum_{j = 0}^{n - i} j = i(n - i + 1) + \frac{(n-i)(n-i+1)}{2} $$ Note that this directly gives you the final answer.

Answer 2

$(k+1)i$ comes from

$i+0+i+i+i+...i$ which contains $(k+1)$ terms. Your sum is the difference

$$\sum_{j=1}^{i+k}j-\sum_{j=1}^{i-1}j=$$

$$\frac{(i+k)(i+k+1)}{2}-\frac{i(i-1)}{2}=$$

$$\frac{k(k+2i+1)+2i}{2}=i(k+1)+\frac{k(k+1)}{2}.$$