Give the defenses i , and ii ,and give and answer to c

Defense i:

Step 1: Assume the statement for $n = k$.

$P(k): \sum_{i=1}^k i = \frac{k(k+1)}{2}$

Step 2: Prove for $n = k+1$.

$\sum_{i=1}^{k+1} i = \sum_{i=1}^k i + (k+1)$

Substitute $P(k)$:

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

Thus, $P(k+1)$ holds.

Defense ii:

Step 1: Base case $n=1$.

$\sum_{i=1}^1 i = 1 = \frac{1 \cdot 2}{2}$

Holds.

Part c:

Step 1: Compute $\sum_{i=1}^{10} i$.

Formula: $\frac{n(n+1)}{2}$, $n=10$.

$\frac{10 \cdot 11}{2} = 55$

55