Let S(K) = 1 + 3 + 5 + ... + (2K - 1) = 3 + K². Then which of the following is true (a) Principle of

Q: Wait, why is $S(1)$ false? Isn't the sum of the first odd number just 1?

You are absolutely correct! The sum of the first odd number is indeed 1. However, the formula provided in the problem states $S(K) = 3 + K^2$. If you plug in $K=1$, you get $3 + 1^2 = 4$. Since $1 \neq 4$, the formula fails at the very first step. Never assume a formula is correct just because it looks like a standard summation; always test the base case with the specific expression given.

Q: How can $S(K) \implies S(K+1)$ be true if $S(K)$ is false? This feels like a contradiction.

This is a brilliant question that touches on the heart of formal logic! In logic, an implication $P \implies Q$ is a statement about the *relationship* between $P$ and $Q$. It asserts that *if* $P$ were true, then $Q$ would necessarily follow. It does not require $P$ to be true in reality. We have proven that the algebraic structure of the formula is consistent, even if the formula itself is incorrect for the base case.

Q: If the implication is true, can we still use induction to prove the formula?

No, we cannot. The Principle of Mathematical Induction requires two distinct pillars: a true base case AND a valid inductive step. Think of it like a bridge: the base case is the anchor on one side, and the inductive step is the span of the bridge. If the anchor is missing, the bridge cannot stand, no matter how perfectly the span is constructed. Since $S(1)$ is false, the induction fails.

Question

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

Options:

(A) Principle of mathematical induction can be used to prove the formula

(B) $S(K) \Rightarrow S(K + 1)$

(C) $S(K) 
ot\Rightarrow S(K + 1)$

(D) $S(1)$ is correct

Sigma Percentile · Accepted Answer

### The Domino Paradox: A Lesson in Logical Rigor

Welcome, future engineers! Today, we are going to dissect a problem that is less about complex calculus and more about the fundamental architecture of mathematical logic.

We are looking at the Principle of Mathematical Induction (PMI), a tool that is as powerful as it is delicate. The problem presents us with a statement $S(K) = 1 + 3 + 5 + ... + (2K - 1) = 3 + K^2$ and asks us to evaluate its validity.

### Phase 1: The Reality Check (The Base Case)

Imagine you are standing before a long line of dominoes. The Principle of Mathematical Induction is our way of ensuring that if we tip the first one, the entire line will fall.

But what if the first domino is glued to the floor? That is exactly what we are checking with the base case, $S(1)$.

We are given the statement $S(K) = 1 + 3 + 5 + ... + (2K - 1) = 3 + K^2$. To test the base case, we set $K = 1$.

On the left-hand side (LHS), we have the sum of the first odd number, which is simply $1$. On the right-hand side (RHS), we substitute $K = 1$ into the formula: $3 + (1)^2 = 4$.

Comparing the two, we see that $1 
eq 4$. The base case is false! The first domino is indeed glued to the floor. This immediately tells us that we cannot use induction to prove this formula, as the starting point is invalid.

### Phase 2: The Algebraic Dance (The Inductive Step)

Even though the base case failed, the question asks us to evaluate the implication $S(K) \implies S(K+1)$. This is where students often get confused.

They think, "If the formula is wrong, how can the step be right?" But remember, logic is cold and calculating. We are not asking if the formula is true; we are asking if the **transition** from $K$ to $K+1$ is algebraically sound.

Let us assume $S(K)$ is true for some arbitrary integer $K$. Our assumption is:
$$1 + 3 + 5 + ... + (2K - 1) = 3 + K^2$$

To reach $S(K+1)$, we need to add the $(K+1)^{th}$ term to both sides. The $(K+1)^{th}$ term is found by substituting $K+1$ into the general term $(2K - 1)$, which gives us $2(K+1) - 1 = 2K + 1$.

Now, we add this term to both sides of our assumed equation:
$$LHS = (3 + K^2) + (2K + 1)$$

This is where the magic happens. Let us rearrange the terms on the right side:
$$LHS = 3 + (K^2 + 2K + 1)$$

Look at that expression inside the parentheses. It is a perfect square! Since $K^2 + 2K + 1 = (K+1)^2$, our equation becomes:
$$LHS = 3 + (K+1)^2$$

This is exactly the form of $S(K+1)$. We have successfully shown that if $S(K)$ were true, $S(K+1)$ would necessarily follow. The algebraic bridge is perfectly built, even if the bridge leads to nowhere because the starting point is missing.

### Conclusion: The Beauty of Logic

We have arrived at a fascinating conclusion. The logical implication $S(K) \implies S(K+1)$ is valid, but the Principle of Mathematical Induction cannot be used to prove the formula because the base case $S(1)$ is false.

This problem is a masterclass in distinguishing between the validity of a logical step and the truth of a statement. In your JEE journey, always remember: **rigor is not just about calculating the right answer; it is about understanding the conditions under which your tools are allowed to operate.**

Keep questioning, keep calculating, and keep falling in love with the logic behind the math!

Animated Solution for Mathematics - Sequence and Series: Let $S (K) = 1 + 3 + 5 + ... + (2 K - 1) = 3 + K^{2}$ . Then which of the following is true

Select Answer:

Visualized Solution

Checking the Base Case $S (1)$

Evaluating the Base Case

Impact on Mathematical Induction

The Inductive Hypothesis $S (K)$

Transitioning to $S (K + 1)$

Evaluating $S (K + 1)$

Conclusion on the Implication

The Sigma Insight: Sum of Special Series

The Domino Paradox

Phase 1

Phase 2

Conclusion

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S(K)=1+3+5+...+(2K−1)=3+K2. Then which of the following is true

Select Answer:

Visualized Solution

Checking the Base Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S(1)

Evaluating the Base Case

Impact on Mathematical Induction

The Inductive Hypothesis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S(K)

Transitioning to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S(K+1)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S(K+1)