NIMCET Binomial Theorem Previous Year Question Paper and Solution with PDF

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NIMCET Binomial Theorem PYQ

Qus : 1 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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NIMCET Previous Year PYQ 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 : 2 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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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Qus : 3 NIMCET PYQ

If $(1 + x - 2x^2)^6 = 1 + a_1 x + a_2 x^2 + \ldots + a_{12} x^{12}$, then the value of $a_2 + a_4 + a_6 + \ldots + a_{12}$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

Even–power coefficient sum = $\dfrac{f(1) + f(-1)}{2}$

$f(1) = (1 + 1 - 2)^6 = 0^6 = 0$

$f(-1) = (1 - 1 - 2)^6 = (-2)^6 = 64$

Required sum $= \dfrac{0 + 64}{2} = 32$

Qus : 4 NIMCET PYQ

The sum $^{20}C_8 + ^{20}C_9 + ^{21}C_{10} + ^{22}C_{11} - ^{23}C_{11}$

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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

Solution

Use Pascal identity: $^{n}C_r + ^{n}C_{r+1} = ^{n+1}C_{r+1}$

So, $^{20}C_8 + ^{20}C_9 = ^{21}C_9$

Then, $^{21}C_9 + ^{21}C_{10} = ^{22}C_{10}$

Next, $^{22}C_{10} + ^{22}C_{11} = ^{23}C_{11}$

Thus the expression becomes:

$^{23}C_{11} - ^{23}C_{11} = 0$

$\therefore$ the answer is 0.

Qus : 5 NIMCET PYQ

If $\displaystyle \sum_{K=0}^{2n}(-1)^K\binom{2n}{K}^2 = A$, find $\displaystyle \sum_{K=0}^{2n}(-1)^K(K-2n)\binom{2n}{K}^2$.

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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

$\displaystyle \sum (-1)^K (K-n)\binom{2n}{K}^2 = 0$ Shifting to $(K-2n)$ keeps symmetry ⇒ still $0$.

Qus : 6 NIMCET PYQ

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

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Qus : 7 NIMCET PYQ

If $a$ and $b$ are the greatest values of $^{2n}C_r$ and $^{2n-1}C_r$ respectively, then:

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Qus : 8 NIMCET PYQ

In the expression $(x+1)(x+4)(x+9)(x+16)\cdots(x+400)$ the coefficient of $x^{19}$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

Coefficient of $x^{19}$ equals sum of constants: $1+4+9+\cdots+400$ This is sum of squares from $1^2$ to $20^2$: $\frac{20(21)(41)}{6}=2870$ Answer: $\boxed{2870}$

Qus : 9 NIMCET PYQ

If $(1+x)^n = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$, then $\left(1+\frac{a_1}{a_0}\right)\left(1+\frac{a_2}{a_1}\right)\left(1+\frac{a_3}{a_2}\right)\dots\left(1+\frac{a_n}{a_{n-1}}\right)$

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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

Solution

Using $a_k = \binom{n}{k}$: $1 + \frac{a_k}{a_{k-1}} = \frac{n+1}{k}$ Product = $\frac{(n+1)^n}{n!}$

Qus : 10 NIMCET PYQ

The coefficient of $x^n$ in the expansion of $(1 - 2x + 3x^2 - 4x^3 + \cdots \text{ to } \infty)^{-n}$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 11 NIMCET PYQ

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

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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

Qus : 12 NIMCET PYQ

The coefficient of $x^{10}$ in the expansion of $\left(x^2+\frac{1}{x}\right)^{12}+\left(x+\frac{1}{x^2}\right)^{12}$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

First expression is:

$\left(x^2+\frac{1}{x}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x^2)^{12-k}\left(\frac{1}{x}\right)^k$

$T_{k+1}={}^{12}C_k x^{24-2k-k}$

$T_{k+1}={}^{12}C_k x^{24-3k}$

For coefficient of $x^{10}$,

$24-3k=10$

$3k=14$

$k=\frac{14}{3}$

This is not an integer, so $x^{10}$ term is not present in the first expression.

Now second expression is:

$\left(x+\frac{1}{x^2}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x)^{12-k}\left(\frac{1}{x^2}\right)^k$

$T_{k+1}={}^{12}C_k x^{12-k-2k}$

$T_{k+1}={}^{12}C_k x^{12-3k}$

For coefficient of $x^{10}$,

$12-3k=10$

$3k=2$

$k=\frac{2}{3}$

This is also not an integer, so $x^{10}$ term is not present in the second expression.

Therefore, the coefficient of $x^{10}$ is $0$.

Qus : 13 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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NIMCET Previous Year Questions (PYQs)

NIMCET Binomial Theorem PYQ

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

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