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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Binomial Theorem PYQ


MCA NIMCET PYQ
The coefficient of $x^{50}$ in the expression of ${(1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + ...... + 1001x^{1000}}$





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2023 PYQ

Solution

Easiest Method — Coefficient of \( x^{50} \)

Given Expression:

\[ (1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + \cdots + 1001x^{1000} \]

This follows a known identity that simplifies the full expression to:

\[ f(x) = (1 + x)^{1002} \]

Now: The coefficient of \( x^{50} \) in \( f(x) \) is:

\[ \boxed{\binom{1002}{50}} \]

✅ Final Answer:   \( \boxed{\binom{1002}{50}} \)


MCA NIMCET PYQ
Which of the following number is the coefficient of $x^{100}$ in the expansion of $\log _e\Bigg{(}\frac{1+x}{1+{x}^2}\Bigg{)},\, |x|{\lt}1$ ?





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2023 PYQ

Solution


MCA NIMCET PYQ
If (1 + x – 2x2)= 1 + a1x + a2x+ ... + a12x12, then the value a+ a+ a+ ... + a12





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2019 PYQ

Solution



MCA NIMCET PYQ
The coefficient of  in the expansion of is





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2018 PYQ

Solution


MCA NIMCET PYQ
If (1 - x + x)n = a + a1x + a2x2 + ... + a2nx2n , then a0 + a2 + a4 + ... + a2n is





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2017 PYQ

Solution


MCA NIMCET PYQ
If $x$ is so small that $x^{2}$ and higher powers of $x$ can be neglected, then $\frac{(9+2x)^{1/2}(3+4x)}{(1-x)^{1/5}}$ is approximately equal to





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2014 PYQ

Solution



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