Let S(k) = 1 + 3 + 5 + dots + (2k - 1) = 3 + | vedclass

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(C) Given $S(k) = 1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$. For $k = 1$,$S(1) \Rightarrow 1 = 3 + 1^2 = 4$,which is false. For $k = 2$,$S(2) \Rightarro

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$S(k) \Rightarrow S(k + 1)$

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(C) Given $S(k) = 1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$. For $k = 1$,$S(1) \Rightarrow 1 = 3 + 1^2 = 4$,which is false. For $k = 2$,$S(2) \Rightarrow 1 + 3 = 3 + 2^2 = 7$,which is false. Now,assume $S(k)$ is true,i.e.,$1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$. Adding $(2(k + 1) - 1) = 2k + 1$ to both sides: $1 + 3 + 5 + \dots + (2k - 1) + (2k + 1) = 3 + k^2 + 2k + 1 = 3 + (k + 1)^2$. This shows that $S(k) \Rightarrow S(k + 1)$ is true,even though the base case $S(1)$ is false.

Let $S(k) = 1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$. Then which of the following is true?

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