MCA NIMCET Binomial Theorem Previous Year Question Paper and Solution with PDF

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Binomial Theorem PYQ

Qus : 1MCA NIMCET PYQ 2023

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: $\boxed{\binom{1002}{50}}$

Qus : 2MCA NIMCET PYQ 2023

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

Qus : 3MCA NIMCET PYQ 2019

If (1 + x – 2x²)⁶= 1 + a₁x + a₂x²+ ... + a₁₂x¹², then the value a₂+ a₄+ a₆+ ... + a₁₂

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 4MCA NIMCET PYQ 2019

Not Available

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 5MCA NIMCET PYQ 2018

The coefficient of

in the expansion of

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 6MCA NIMCET PYQ 2017

If (1 - x + x²)ⁿ = a + a₁x + a₂x² + ... + a_2nx²ⁿ , then a₀ + a₂ + a₄ + ... + a_2n is

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 7MCA NIMCET PYQ 2014

$x$ 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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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Binomial Theorem PYQ

Solution

Easiest Method — Coefficient of x^{50} x^{50}

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Easiest Method — Coefficient of $x^{50}$