Answer
a)P(1) is true ,P(1) is “$1^2$
b)derived in step by step
c)derived in step by step
d)derived in step by step
Work Step by Step
a) P(1) is “$1^2$ = $\frac{1 ·(1+1) · (2 · 1+1)}{6}$ .”
P(1) is true because
$1^2$ = 1 and$\frac{ 1 · (1+1) ·(2+1)}{6}$= $\frac{2 · 3}{6}$= 1 also.
b. P(k) is “$1^2$ + $2^2$ +· · ·+$k^2$ = $\frac{k(k+1)(2k+1)}{6}$ .”
c. P(k + 1) is “$1^2$ + $2^2$ +· · ·+$(k + 1)^2$=$\frac{(k+1)((k+1)+1)(2 · (k+1)+1)}{6}$.”
d. Must show: If for some integer k ≥ 1,
$1^2$ + $2^2$ +· · ·+$k^2$ = $\frac{k(k+1)(2k+1)}{6}$ ,
then
$1^2 + 2^2 +· · ·+(k + 1)^2$ =$\frac{ (k+1)[(k+1)+1][(2(k+1)+1)]}{6}$ .
