NIMCET 2027 Binomial Theorem Previous Year Questions (PYQs) – Page 1 of 2
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Showing 13 questions
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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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: \( \boxed{\binom{1002}{50}} \)
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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Solution
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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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 $= \dfrac{0 + 64}{2} = 32$
The sum
$^{20}C_8 + ^{20}C_9 + ^{21}C_{10} + ^{22}C_{11} - ^{23}C_{11}$
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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.
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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Solution
$\displaystyle \sum (-1)^K (K-n)\binom{2n}{K}^2 = 0$
Shifting to $(K-2n)$ keeps symmetry ⇒ still $0$.
If (1 + x – 2x2)6 = 1 + a1x + a2x2 + ... + a12x12,
then the value a2 + a4 + a6 + ... + a12
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Solution
If \(a\) and \(b\) are the greatest values of \(^{2n}C_r\) and \(^{2n-1}C_r\) respectively, then:
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Solution
In the expression
$(x+1)(x+4)(x+9)(x+16)\cdots(x+400)$
the coefficient of $x^{19}$ is
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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: $\boxed{2870}$
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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Solution
Using $a_k = \binom{n}{k}$:
$1 + \frac{a_k}{a_{k-1}} = \frac{n+1}{k}$
Product = $\frac{(n+1)^n}{n!}$
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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Solution
NIMCET 2027
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