If p be a natural number then prove...

MathematicsBinomial Expansion for Positive Integral IndexJEE Advanced 1984Easy

If $p$ be a natural number then prove that $p^{n + 1} + (p + 1)^{2 n - 1}$ is divisible by $p^{2} + p + 1$ for every positive integer $n$ .

Visualized Solution (Hindi)

Define the Proposition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (n)

Let the given expression be denoted by $P (n)$ .
We define $P (n) = p^{n + 1} + (p + 1)^{2 n - 1}$ .
Our goal is to prove that $P (n)$ is divisible by $p^{2} + p + 1$ for all $n \in Z^{+}$ .
We will use the Principle of Mathematical Induction (PMI) for this proof.

Verify the Base Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n = 1

Substitute $n = 1$ into the expression $P (n)$ .
Calculation: $P (1) = p^{1 + 1} + (p + 1)^{2 (1) - 1}$
Simplification: $P (1) = p^{2} + (p + 1)^{1} = p^{2} + p + 1$
Since $p^{2} + p + 1$ is divisible by itself, the base case $P (1)$ is true.

State the Inductive Hypothesis

Assume that $P (k)$ is true for some positive integer $k$ .
This means $P (k) = p^{k + 1} + (p + 1)^{2 k - 1} = (p^{2} + p + 1) m$ for some integer $m$ .
From this, we can express $p^{k + 1}$ as:
Equation: $p^{k + 1} = (p^{2} + p + 1) m - (p + 1)^{2 k - 1}$

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n = k + 1

Now, consider the expression for $n = k + 1$ .
$P (k + 1) = p^{(k + 1) + 1} + (p + 1)^{2 (k + 1) - 1}$
Simplifying the exponents: $P (k + 1) = p^{k + 2} + (p + 1)^{2 k + 1}$

Decompose the Terms

Break down the powers to match the terms in $P (k)$ .
$P (k + 1) = p \cdot p^{k + 1} + (p + 1)^{2} (p + 1)^{2 k - 1}$
This decomposition allows us to use the substitution from our inductive hypothesis.

Substitute the Hypothesis

Substitute $p^{k + 1} = (p^{2} + p + 1) m - (p + 1)^{2 k - 1}$ into the equation.
$P (k + 1) = p [(p^{2} + p + 1) m - (p + 1)^{2 k - 1}] + (p + 1)^{2} (p + 1)^{2 k - 1}$
Now we have the entire expression in terms of $m$ and $(p + 1)$ .

Factorize and Simplify

Expand the expression: $P (k + 1) = p (p^{2} + p + 1) m - p (p + 1)^{2 k - 1} + (p^{2} + 2 p + 1) (p + 1)^{2 k - 1}$
Group the terms with $(p + 1)^{2 k - 1}$ : $P (k + 1) = p (p^{2} + p + 1) m + (p^{2} + 2 p + 1 - p) (p + 1)^{2 k - 1}$
Simplify the bracket: $P (k + 1) = p (p^{2} + p + 1) m + (p^{2} + p + 1) (p + 1)^{2 k - 1}$
Factor out the common divisor: $P (k + 1) = (p^{2} + p + 1) [p m + (p + 1)^{2 k - 1}]$

Conclusion by Induction

Since $P (k + 1)$ is a multiple of $(p^{2} + p + 1)$ , it is divisible by $(p^{2} + p + 1)$ .
By the Principle of Mathematical Induction, $P (n)$ is true for all $n \in Z^{+}$ .
Key Takeaway: Decomposition of exponents is the primary tool for solving induction-based divisibility problems.
Next Challenge: Try proving divisibility for $3^{2 n + 2} - 8 n - 9$ by $64$ using a similar approach.

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