Explanation

(b)
$S(k) = 1 + 3 + 5$….$+ (2k - 1) = 3 + k^{2}$
Put $k = 1$ in both sides, we get
LHS $= 1$ and RHS$= 3 + 1 = 4$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS \neq RHS$
Put $(k + 1)$ in both sides in the place of $k$, we get
$\ \ \ \ \ LHS = 1 + 3 + 5$….$+ (2k - 1) + (2k + 1)$
$\ \ \ \ RHS = 3 + (k + {1)}^{2} = 3 + k^{2} + 2k + 1$
Let $LHS = RHS$
Then,$1 + 3 + 5$….$+ (2k - 1) + (2k + 1)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3 + k^{2} + 2k + 1$
$\Longrightarrow 1 + 3 + 5 +$…$+ (2k - 1) = 3 + k^{2}$
If $S(k)$ is true, then $S(k + 1)$ is also true.
Hence, $S(k) \Longrightarrow S(k + 1)$