Prove that 7²ⁿ + 2³ⁿ⁻³(3ⁿ⁻¹) is...

MathematicsBinomial Expansion for Positive Integral IndexJEE Advanced 1982Easy

Prove that $7^{2 n} + 2^{3 n - 3} (3^{n - 1})$ is divisible by 25 for any natural number $n$ .

Visualized Solution (English)

Simplifying the Expression .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 $P (n) = 7^{2 n} + 2^{3 n - 3} \times 3^{n - 1}$
Simplify the first term: $7^{2 n} = (7^{2})^{n} = 4 9^{n}$
Simplify the second term: $2^{3 n - 3} \times 3^{n - 1} = (2^{3})^{n - 1} \times 3^{n - 1} = 8^{n - 1} \times 3^{n - 1} = (8 \times 3)^{n - 1} = 2 4^{n - 1}$
The simplified expression is $P (n) = 4 9^{n} + 2 4^{n - 1}$

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

For $n = 1$ : $P (1) = 4 9^{1} + 2 4^{1 - 1}$
Evaluate the powers: $P (1) = 49 + 2 4^{0} = 49 + 1$
Final sum: $P (1) = 50$
Check divisibility: $50 = 2 \times 25$ , which is divisible by $25$ .
The base case is True.

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

Assume the statement is true for $n = k$ .
Let $P (k) = 4 9^{k} + 2 4^{k - 1} = 25 m$ for some integer $m$ .
Rearrange for later use: $4 9^{k} = 25 m - 2 4^{k - 1}$

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

We need to prove $P (k + 1)$ is divisible by $25$ .
$P (k + 1) = 4 9^{k + 1} + 2 4^{(k + 1) - 1}$
Simplify the exponent: $P (k + 1) = 4 9^{k + 1} + 2 4^{k}$
Break down the first term: $P (k + 1) = 49 \times 4 9^{k} + 2 4^{k}$

Substitution from Hypothesis

Substitute $4 9^{k} = 25 m - 2 4^{k - 1}$ into $P (k + 1)$ :
$P (k + 1) = 49 (25 m - 2 4^{k - 1}) + 2 4^{k}$

Algebraic Simplification

Expand the expression: $P (k + 1) = 49 \times 25 m - 49 \times 2 4^{k - 1} + 2 4^{k}$
Rewrite $2 4^{k}$ as $24 \times 2 4^{k - 1}$
$P (k + 1) = 49 \times 25 m - 49 \times 2 4^{k - 1} + 24 \times 2 4^{k - 1}$
Combine like terms: $P (k + 1) = 49 \times 25 m - 25 \times 2 4^{k - 1}$

Conclusion

Factor out $25$ : $P (k + 1) = 25 (49 m - 2 4^{k - 1})$
Since $49 m - 2 4^{k - 1}$ is an integer, $P (k + 1)$ is divisible by $25$ .
By the Principle of Mathematical Induction, $P (n)$ is divisible by $25$ for all $n \in N$ .

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Prove that $7^{2 n} + 2^{3 n - 3} (3^{n - 1})$ is divisible by 25 for any natural number $n$ .

Visualized Solution (English)

Simplifying the Expression $P (n)$

Base Case: $n = 1$

Inductive Hypothesis: $n = k$

Inductive Step: $n = k + 1$

Substitution from Hypothesis

Algebraic Simplification

Conclusion

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that 72n+23n−3(3n−1) is divisible by 25 for any natural number n.

Visualized Solution (English)

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

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

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

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

Substitution from Hypothesis

Algebraic Simplification

Conclusion

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Prove that $7^{2 n} + 2^{3 n - 3} (3^{n - 1})$ is divisible by 25 for any natural number $n$ .

Simplifying the Expression $P (n)$

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Inductive Step: $n = k + 1$

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Use mathematical Induction to prove : If $n$ is any odd positive integer, then $n (n^{2} - 1)$ is divisible by 24.

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