Let n ≥ 5 be an integer. If 9n – 8n – 1 = 64α and 6n – 5n – 1 = 25β, then α – β is equal to
\(1 + ^nC_2 (8-5) + ^nC_3(8² - 5²) + ... + ^nC_n ( 8^{n-1} - 5^{n-1} )\)
\(1 + ^nC_3 (8-5) + ^nC_4(8² - 5²) + ... + ^nC_n ( 8^{n-2} - 5^{n-2} )\)
\(^nC_3 (8-5) + ^nC_4 ( 8²-5²) + ... + ^nC_n ( 8^{n-2} - 5^{n-2} )\)
\(^nC_4 (8-5) + ^nC_5 ( 8²-5²) + ... + ^nC_n ( 8^{n-3} - 5^{n-3} )\)
The correct answer is (C) : \(^nC_3 (8-5) + ^nC_4 ( 8²-5²) + ... + ^nC_n ( 8^{n-2} - 5^{n-2} )\)
(1 + 8)n – 8n – 1 = 64α
\(⇒ 1 + 8n + ^nC_28^2 + ^nC_38^3 + ..... +^nC_n8^n - 8n-1 = 64α\)
\(⇒ α = ^nC_2 + ^nC_38 + ^nC_48^2 + ..... + ^nC_n8^{n-2} ...... (i)\)
Similarly
(1+5)n - 5n-1=25β
\(⇒ 1 + 5n + ^nC_25^2 + ^nC_35^3 + ..... + ^nC_n5^n - 5n - 1 = 25β\)
\(⇒ β = ^nC_2 + ^nC_3.5 + ^nC_4.5^2 + ...... + ^nC_n 5^{n-2} ..... (ii)\)
\(α - β = ^nC_3(8-5) + ^nC_4 (8^2-5^2) + ..... + ^nC_n(8^{n-2} - 5^{n-2})\)
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
