Prove by induction that ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2}∑i=1ni=2n(n+1). Which statement is part of the inductive step?
Assume ∑i=1ki=k(k+1)2\sum_{i=1}^{k} i = \frac{k(k+1)}{2}∑i=1ki=2k(k+1) for some k≥1k \geq 1k≥1, then show ∑i=1k+1i=(k+1)(k+2)2\sum_{i=1}^{k+1} i = \frac{(k+1)(k+2)}{2}∑i=1k+1i=2(k+1)(k+2).
Verify that ∑i=11i=1=1⋅22\sum_{i=1}^{1} i = 1 = \frac{1 \cdot 2}{2}∑i=11i=1=21⋅2.
Show that ∑i=1ni≤n(n+1)2\sum_{i=1}^{n} i \leq \frac{n(n+1)}{2}∑i=1ni≤2n(n+1) for all nnn.
Compute ∑i=1100i\sum_{i=1}^{100} i∑i=1100i directly.