Let R = (5√5 + 11)²ⁿ⁺¹ and f = R - [R],...

MathematicsBinomial Expansion for Positive Integral IndexJEE Advanced 1988Moderate

Let $R = (55 + 11)^{2 n + 1}$ and $f = R - [R]$ , where $[]$ denotes the greatest integer function. Prove that $R f = 4^{2 n + 1}$ .

Visualized Solution (English)

Define the Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f

Given: $R = (55 + 11)^{2 n + 1}$
Given: $f = R - [R]$ , where $[R]$ is the greatest integer part.
To Prove: $R f = 4^{2 n + 1}$

Introduce the Conjugate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R^{'}

Let $R^{'} = (55 - 11)^{2 n + 1}$
This is the conjugate of the original expression $R$ .

Establish the Bounds for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R^{'}

Since $11 < 55 < 12$ , we have $0 < 55 - 11 < 1$ .
Raising a value between $0$ and $1$ to any positive power $2 n + 1$ results in a value between $0$ and $1$ .
Therefore, $0 < R^{'} < 1$ .

Binomial Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R^{'}

$R = (55 + 11)^{2 n + 1} = \sum_{k = 0}^{2 n + 1}^{2 n + 1} C_{k} (55)^{2 n + 1 - k} (11)^{k}$
$R^{'} = (55 - 11)^{2 n + 1} = \sum_{k = 0}^{2 n + 1}^{2 n + 1} C_{k} (55)^{2 n + 1 - k} (- 11)^{k}$

Subtracting the Expansions

$R - R^{'} = 2 [^{2 n + 1} C_{1} (55)^{2 n} \cdot 11 +^{2 n + 1} C_{3} (55)^{2 n - 2} \cdot 1 1^{3} + \dots]$
Since $(55)^{2} = 125$ is an integer, the term in the bracket is an integer.
Thus, $R - R^{'} = Even Integer$ .

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R, f, and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R^{'}

Substitute $R = [R] + f$ into $R - R^{'} = I$ (where $I$ is an even integer).
$[R] + f - R^{'} = I \Rightarrow f - R^{'} = I - [R]$
Since $I$ and $[R]$ are integers, $f - R^{'}$ must be an integer.

Proving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f = R^{'}

We have $0 < f < 1$ and $0 < R^{'} < 1$ .
Subtracting these inequalities: $- 1 < f - R^{'} < 1$ .
The only integer in the interval $(- 1, 1)$ is $0$ .
Therefore, $f - R^{'} = 0 \Rightarrow f = R^{'}$ .

Final Product Calculation

$R f = R \cdot R^{'} = (55 + 11)^{2 n + 1} (55 - 11)^{2 n + 1}$
$R f = [(55 + 11) (55 - 11)]^{2 n + 1}$
$R f = [(55)^{2} - 1 1^{2}]^{2 n + 1} = [125 - 121]^{2 n + 1}$
$R f = 4^{2 n + 1}$

Key Takeaway and Summary

Key Takeaway: Use the conjugate $R^{'}$ to eliminate irrational terms in fractional part problems.
Logic Check: Always verify the bounds of the conjugate ( $0 < R^{'} < 1$ ).
Next Challenge: Try solving for $R = (33 + 5)^{2 n}$ and find the relation between $R$ and $f$ .

00:00 / 00:00

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Let $R = (55 + 11)^{2 n + 1}$ and $f = R - [R]$ , where $[]$ denotes the greatest integer function. Prove that $R f = 4^{2 n + 1}$ .

Visualized Solution (English)

Define the Expression $R$ and $f$

Introduce the Conjugate $R^{'}$

Establish the Bounds for $R^{'}$

Binomial Expansion of $R$ and $R^{'}$

Subtracting the Expansions

Relating $R, f,$ and $R^{'}$

Proving $f = R^{'}$

Final Product Calculation

Key Takeaway and Summary

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R=(55​+11)2n+1 and f=R−[R], where [ ] denotes the greatest integer function. Prove that Rf=42n+1.

Visualized Solution (English)

Define the Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f

Introduce the Conjugate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R′

Establish the Bounds for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R′

Binomial Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R′

Subtracting the Expansions

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R,f, and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R′

Proving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f=R′

Final Product Calculation

Key Takeaway and Summary

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Let $R = (55 + 11)^{2 n + 1}$ and $f = R - [R]$ , where $[]$ denotes the greatest integer function. Prove that $R f = 4^{2 n + 1}$ .

Define the Expression $R$ and $f$

Introduce the Conjugate $R^{'}$

Establish the Bounds for $R^{'}$

Binomial Expansion of $R$ and $R^{'}$

Relating $R, f,$ and $R^{'}$

Proving $f = R^{'}$

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Given $n^{4} < 1 0^{n}$ for a fixed positive integer $n \geq 2$ , prove that $(n + 1)^{4} < 1 0^{n + 1}$ .

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

Prove by mathematical induction that $\frac{( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} \leq \frac{1}{( 3 n + 1 ) ^{1/2}}$ for all positive integers $n$ .

Prove by permutation or otherwise $\frac{( n ^{2} )!}{( n ! ) ^{n}}$ is an integer ( $n \in I^{+}$ ).

Show that $\int_{0}^{nπ + v} ∣ sin x ∣ d x = 2 n + 1 - cos v$ where $n$ is a positive integer and $0 \leq v < π$ .

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

Let $R = (55 + 11)^{2 n + 1}$ and $f = R - [R]$ , where $[]$ denotes the greatest integer function. Prove that $R f = 4^{2 n + 1}$ .

Prove that $\frac{n ^{7}}{7} + \frac{n ^{5}}{5} + \frac{2 n ^{3}}{3} - \frac{n}{105}$ is an integer for every positive integer $n$ .

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Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

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