Let be an integer. If and , then is:

Explanation: Solution We will solve this using the Binomial Theorem : Step 1: Find the value of Given . We can write as : Substituting this into the equation for : The and terms cancel out: Divide both sides by (which is ): Step 2: Find the value of Note: In the image, the second equation is . Usually, in these problems, it follows the pattern . Let's solve for using the standard binomial expansion . Substituting this: The and terms cancel out: Divide both sides by (which is ): Step 3: Calculate Subtract the expression for from : Final Answer The result matches Option (C) .

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JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

Let $n \ge 5$ be an integer. If $9^n - 8n - 1 = 64\alpha$ and $6^n - 5n - 1 = 25\beta$ , then $\alpha - \beta$ is:

Choose the correct answer:

Explanation

Solution

We will solve this using the Binomial Theorem:

$(1 + x)^n = 1 + nx + {}^nC_2x^2 + {}^nC_3x^3 + \dots + {}^nC_nx^n$

Step 1: Find the value of $\alpha$

Given $64\alpha = 9^n - 8n - 1$ .

We can write $9^n$ as $(1 + 8)^n$ :

$(1 + 8)^n = 1 + n(8) + {}^nC_2(8)^2 + {}^nC_3(8)^3 + \dots + {}^nC_n(8)^n$

Substituting this into the equation for $\alpha$ :

$64\alpha = [1 + 8n + {}^nC_2(8^2) + {}^nC_3(8^3) + \dots + {}^nC_n(8^n)] - 8n - 1$

The $1$ and $8n$ terms cancel out:

$64\alpha = {}^nC_2(8^2) + {}^nC_3(8^3) + \dots + {}^nC_n(8^n)$

Divide both sides by $64$ (which is $8^2$ ):

$\alpha = {}^nC_2 + {}^nC_3(8) + {}^nC_4(8^2) + \dots + {}^nC_n(8^{n-2})$

Step 2: Find the value of $\beta$

Note: In the image, the second equation is $6^n - 5n = 25\beta$ . Usually, in these problems, it follows the pattern $6^n - 5n - 1 = 25\beta$ . Let's solve for $\beta$ using the standard binomial expansion $6^n = (1+5)^n$ .

$64\beta = (1 + 5)^n - 5n - 1$

$(1 + 5)^n = 1 + n(5) + {}^nC_2(5^2) + {}^nC_3(5^3) + \dots + {}^nC_n(5^n)$

Substituting this:

$25\beta = [1 + 5n + {}^nC_2(5^2) + {}^nC_3(5^3) + \dots + {}^nC_n(5^n)] - 5n - 1$

The $1$ and $5n$ terms cancel out:

$25\beta = {}^nC_2(5^2) + {}^nC_3(5^3) + \dots + {}^nC_n(5^n)$

Divide both sides by $25$ (which is $5^2$ ):

$\beta = {}^nC_2 + {}^nC_3(5) + {}^nC_4(5^2) + \dots + {}^nC_n(5^{n-2})$

Step 3: Calculate $\alpha - \beta$

Subtract the expression for $\beta$ from $\alpha$ :

$\alpha - \beta = ({}^nC_2 - {}^nC_2) + {}^nC_3(8 - 5) + {}^nC_4(8^2 - 5^2) + \dots + {}^nC_n(8^{n-2} - 5^{n-2})$

$\alpha - \beta = {}^nC_3(8 - 5) + {}^nC_4(8^2 - 5^2) + \dots + {}^nC_n(8^{n-2} - 5^{n-2})$

Final Answer

The result matches Option (C).

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