Given:
\( S = x^2(1 + x)^{98} + x^3(1 + x)^{97} + x^4(1 + x)^{96} + \ldots + x^{54}(1 + x)^{46} \)
It is a geometric progression (G.P.).
Therefore,
\( S = x^2(1 + x)^{98} \left[ \frac{\left( \frac{x}{1 + x} \right)^{53} - 1}{\frac{x}{1 + x} - 1} \right] \)
Now, the coefficient of \( x^{70} \) in \( S \) will be:
\( S = x^2(1 + x)^{46} \left[ (1 + x)^{53} - x^{53} \right] \)
The coefficient of \( x^{70} \) in \( S \) is obtained from:
\( S = x^2(1 + x)^{99} - x^{55}(1 + x)^{46} \)
Hence, the required coefficient is:
\( S = {}^{99}C_{68} - {}^{46}C_{15} = {}^{99}C_{p} - {}^{46}C_{q} \)
From this,
\( p = 68, \quad q = 15 \)
Therefore,
\( p + q = 83 \)
\[ x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + …… + x^{54}(1+x)^{46} \]
The coefficient of \(x^{70}\) is:
\[ ^{98}C_{68} + ^{97}C_{67} + ^{96}C_{66} + \cdots \]
Simplify:
\[ ^{47}C_{17} + ^{46}C_{16} \]
Combine terms:
\[ {^{46}}C_{30} + {^{46}}C_{31} + ^{47}C_30 + \cdots \]
Using binomial expansion:
\[ {^{47}}C_{30} + \cdots = ^{99}C_p - ^{46}C_q \]
Possible values of \(p+q\):
\[ p+q = 62, 83, 99, 46 \]
Final Answer:
\[ p+q = 83 \quad \text{Option (4)}. \]
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]