\[ x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + …… + x^{54}(1+x)^{46} \]
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:
\[ p+q = 83 \quad \text{Option (4)}. \]
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is:
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)