Jamia Millia Islamia Binomial Theorem Previous Year Questions (PYQs) – Page 1 of 2
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Showing 13 questions
The total number of terms in the expansion of $(x+a)^{100} + (x-a)^{100}$ after simplification is:
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Solution
$(x+a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}a^r$
$(x-a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}(-a)^r$
Adding, odd powers of $a$ cancel and even powers remain.
Even $r$: total even $r$ from 0 to 100 → 51 terms.
The middle term in the expansion of
$\left(1 + \dfrac{1}{x^2}\right)\!\left(1 + x^2\right)^n$ is —
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Solution
Expand $\left(1 + x^2\right)^n = \sum_{k=0}^{n} {}^{n}C_{k}x^{2k}.$
Multiply by $\left(1 + \dfrac{1}{x^2}\right)$:
$= \sum_{k=0}^{n} {}^{n}C_{k}x^{2k} + \sum_{k=0}^{n} {}^{n}C_{k}x^{2k-2}.$
To find middle term → powers of $x$ that are equal when $2k = 2n - (2k-2)$.
Simplifying gives $k = n.$
So middle term = ${}^{2n}C_{n}.$
The sum of $(n+1)$ terms of the series
$\dfrac{C_0}{2} - \dfrac{C_1}{3} + \dfrac{C_2}{4} - \dfrac{C_3}{5} + \dots$ is
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Solution
Using binomial relation and telescoping pattern, the series reduces to
$\dfrac{1}{(n+1)(n+2)}$
What will be the value of $(102)^5$?
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Solution
Using binomial expansion:
$(100 + 2)^5 = 100^5 + 5(100^4)(2) + 10(100^3)(2^2) + 10(100^2)(2^3) + 5(100)(2^4) + 2^5$
$= 10^{10} + 10^{8}(10) + 10^{6}(40) + 10^{4}(80) + 10^{2}(80) + 32$
$= 11040603032.$
What will be an approximation of $(0.99)^5$ using the first three terms of expansion?
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Solution
Using $(1 - x)^n \approx 1 - nx + \dfrac{n(n-1)}{2}x^2$
For $x = 0.01, n = 5$:
$(1 - 0.01)^5 \approx 1 - 5(0.01) + 10(0.01)^2 = 1 - 0.05 + 0.001 = 0.951.$
The coefficient of the middle term in the binomial expansion of $(1 + ax)^4$ and of $(1 - ax)^6$ is the same, if $a$ is equal to...
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Solution
Middle term of $(1 + ax)^4$ ⇒ $\text{Coefficient} = \binom{4}{2} a^2 = 6a^2.$
Middle term of $(1 - ax)^6$ ⇒ $\text{Coefficient} = \binom{6}{3} (-a)^3 = -20a^3.$
Given equal ⇒ $6a^2 = 20a^3 \Rightarrow a = \dfrac{3}{10}.$
Since signs are opposite ⇒ $a = -\dfrac{3}{10}.$
The coefficient of $x^8 y^{10}$ in $(x + y)^{18}$ is …
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Solution
General term = ${}^{18}C_r x^{18-r} y^r$.
We need $x^8 y^{10} \Rightarrow 18 - r = 8 \Rightarrow r = 10$.
Coefficient = ${}^{18}C_{10}$.
The coefficient of $y$ in the expansion of $\left(y^2+\dfrac{c}{y}\right)^5$ is:
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Solution
General term
$T_k=\binom{5}{k}(y^2)^{5-k}\left(\dfrac{c}{y}\right)^k
=\binom{5}{k}c^k\,y^{10-3k}$.
For power of $y^1$: $10-3k=1\Rightarrow k=3$.
Coefficient $=\binom{5}{3}c^3=10c^3$.
The coefficient of $x^4$ in expansion of $(1 + x + x^2 + x^3)^{11}$ is …
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Solution
We can write generating function $f(x) = (1 + x + x^2 + x^3)^{11}$.
Coefficient of $x^4 =$ number of ways to get power 4 as sum of 11 terms each $0,1,2,3$.
By multinomial expansion, coefficient of $x^4 = {}^{11}C_4 + 10{}^{11}C_3 + 6{}^{11}C_2 + {}^{11}C_1 = 990$.
In the expansion of $(1+x)^{50}$, the sum of coefficients of odd powers of $x$ is:
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Solution
Sum of odd coefficients $=\dfrac{(1+1)^{50}-(1-1)^{50}}{2}
=\dfrac{2^{50}-0}{2}=2^{49}$.
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