Expansion of binomial theorem Questions in English

(A) Given the expression $(1-x)^3(2+x)^6$. Since $x$ is very small,we neglect terms with $x^3$ and higher powers. \\ $(1-x)^3 = 1 - 3x + 3x^2$. \\ $(2+x)^6 = 2^6 + 6 \times 2^5 \times x + \frac{6 \times 5}{2} \times 2^4 \times x^2 = 64 + 192x + 240x^2$. \\ Now,multiply the two expansions: \\ $(1-3x+3x^2)(64+192x+240x^2) = 64 + 192x + 240x^2 - 192x - 576x^2 + 192x^2$ (neglecting $x^3$ and higher). \\ $= 64 + (192-192)x + (240-576+192)x^2 = 64 + 0x - 144x^2$. \\ Comparing with $a+bx+cx^2$,we get $a=64, b=0, c=-144$. \\ Therefore,$a+b+c = 64 + 0 - 144 = -80$.

(D) Given expression: $f(x) = \frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$
Neglecting $x^2$ and higher powers, we use the binomial approximation $(1+u)^n \approx 1+nu$.
$f(x) = (1+\frac{2x}{3})^{-4} \times (4(1+\frac{5x}{4}))^{1/2} \times (9(1+\frac{x}{9}))^{-3/2}$
$f(x) = (1+\frac{2x}{3})^{-4} \times 2(1+\frac{5x}{4})^{1/2} \times \frac{1}{27}(1+\frac{x}{9})^{-3/2}$
$f(x) \approx \frac{2}{27} (1 - 4 \cdot \frac{2x}{3}) (1 + \frac{1}{2} \cdot \frac{5x}{4}) (1 - \frac{3}{2} \cdot \frac{x}{9})$
$f(x) \approx \frac{2}{27} (1 - \frac{8x}{3}) (1 + \frac{5x}{8}) (1 - \frac{x}{6})$
$f(x) \approx \frac{2}{27} (1 - \frac{8x}{3} + \frac{5x}{8} - \frac{x}{6}) = \frac{2}{27} (1 - \frac{64x - 15x + 4x}{24}) = \frac{2}{27} (1 - \frac{53x}{24})$
Substituting $x = \frac{6}{371}$:
$f(\frac{6}{371}) \approx \frac{2}{27} (1 - \frac{53}{24} \cdot \frac{6}{371}) = \frac{2}{27} (1 - \frac{53}{4 \cdot 371}) = \frac{2}{27} (1 - \frac{53}{1484})$
$f(\frac{6}{371}) \approx \frac{2}{27} (\frac{1431}{1484}) = \frac{2}{27} \cdot \frac{53 \cdot 27}{1484} = \frac{106}{1484} = \frac{1}{14}$

(D) We have the expression $\frac{1}{(x-1)^2(x-2)}$.
First,rewrite the expression as $\frac{1}{(-1)^2(1-x)^2(-2)(1-\frac{x}{2})} = -\frac{1}{2}(1-x)^{-2}(1-\frac{x}{2})^{-1}$.
Using the binomial expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$ and $(1-y)^{-1} = 1 + y + y^2 + \dots$ for $|x| < 1$ and $|x/2| < 1$:
$(1-x)^{-2} = 1 + 2x + 3x^2 + \dots$
$(1-\frac{x}{2})^{-1} = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$
Multiplying these series: $-\frac{1}{2}(1 + 2x + \dots)(1 + \frac{x}{2} + \dots) = -\frac{1}{2}(1 + (\frac{1}{2} + 2)x + \dots) = -\frac{1}{2}(1 + \frac{5}{2}x + \dots)$.
The constant term is $-\frac{1}{2} \times 1 = -\frac{1}{2}$.

(C) Let $(2+\sqrt{3})^5 = I + f$,where $I$ is an integer and $0 < f < 1$.
Consider the expression $(2-\sqrt{3})^5 = f'$.
Since $0 < 2-\sqrt{3} < 1$,it follows that $0 < (2-\sqrt{3})^5 < 1$,so $0 < f' < 1$.
Now,consider the sum $S = (2+\sqrt{3})^5 + (2-\sqrt{3})^5$.
Using the binomial expansion,$S = 2 \times [^5C_0 2^5 + ^5C_2 2^3 (\sqrt{3})^2 + ^5C_4 2^1 (\sqrt{3})^4]$.
$S = 2 \times [32 + 10 \times 8 \times 3 + 5 \times 2 \times 9] = 2 \times [32 + 240 + 90] = 2 \times 362 = 724$.
Since $I + f + f' = 724$ and $0 < f + f' < 2$,the sum $f + f'$ must be an integer.
Given $0 < f < 1$ and $0 < f' < 1$,the only possible value for $f + f'$ is $1$.
Therefore,$I + 1 = 724$,which implies $I = 723$.

(D) Given the expression $(1-x-x^2+x^3)^6$.
We can factorize the expression as $[(1-x)-x^2(1-x)]^6 = [(1-x^2)(1-x)]^6 = (1-x^2)^6(1-x)^6$.
Using the binomial expansion $(1+y)^n = \sum_{k=0}^{n} \binom{n}{k} y^k$:
$(1-x^2)^6 = \binom{6}{0} - \binom{6}{1}x^2 + \binom{6}{2}x^4 - \dots = 1 - 6x^2 + 15x^4 - \dots$
$(1-x)^6 = \binom{6}{0} - \binom{6}{1}x + \binom{6}{2}x^2 - \binom{6}{3}x^3 + \binom{6}{4}x^4 - \dots = 1 - 6x + 15x^2 - 20x^3 + 15x^4 - \dots$
To find the coefficient of $x^4$ in the product $(1 - 6x^2 + 15x^4)(1 - 6x + 15x^2 - 20x^3 + 15x^4)$,we multiply terms that result in $x^4$:
$1 \times (15x^4) + (-6x^2) \times (15x^2) + (15x^4) \times (1) = 15 - 90 + 15 = -60$.
Thus,the coefficient of $x^4$ is $-60$.

(A) Given the expression $\frac{1}{(1-5x)(1-4x)}$ for $|x| < \frac{1}{5}$.
Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we have:
$(1-5x)^{-1} = 1 + 5x + 25x^2 + 125x^3 + \dots$
$(1-4x)^{-1} = 1 + 4x + 16x^2 + 64x^3 + \dots$
Multiplying these two series:
$(1 + 5x + 25x^2 + 125x^3 + \dots)(1 + 4x + 16x^2 + 64x^3 + \dots)$
The coefficient of $x^3$ is obtained by collecting terms that result in $x^3$:
$1 \cdot (64x^3) + (5x) \cdot (16x^2) + (25x^2) \cdot (4x) + (125x^3) \cdot 1$
$= 64 + 80 + 100 + 125$
$= 369$

(A) Given $(1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}$.
Using the binomial expansion $(1+y)^n = \sum_{k=0}^n {}^{n}C_k y^k$,let $y = 2x+3x^2$.
$(1+2x+3x^2)^{10} = {}^{10}C_0 + {}^{10}C_1(2x+3x^2) + {}^{10}C_2(2x+3x^2)^2 + \ldots$
$= 1 + 10(2x+3x^2) + 45(4x^2+12x^3+9x^4) + \ldots$
$= 1 + 20x + 30x^2 + 180x^2 + \ldots$
$= 1 + 20x + 210x^2 + \ldots$
Comparing coefficients,we get $a_1 = 20$ and $a_2 = 210$.
Therefore,$\frac{a_2}{a_1} = \frac{210}{20} = 10.5$.

(D) We have,$(1+x+x^2)^n = a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots + a_{2n}x^{2n}$.
On differentiating both sides with respect to $x$,we get:
$n(1+x+x^2)^{n-1}(1+2x) = a_1 + 2a_2x + 3a_3x^2 + \ldots + 2na_{2n}x^{2n-1}$.
Now,putting $x=1$,we get:
$n(1+1+1)^{n-1}(1+2(1)) = a_1 + 2a_2 + 3a_3 + \ldots + 2na_{2n}$.
$n(3)^{n-1}(3) = a_1 + 2a_2 + 3a_3 + \ldots + 2na_{2n}$.
Therefore,$a_1 + 2a_2 + 3a_3 + \ldots + 2na_{2n} = n \cdot 3^n$.

(C) To find the sum of the coefficients in the expansion of a polynomial,we substitute $x = 1$ into the expression.
Given the expansion $\left(1+\frac{x}{2}\right)^{12}$.
Substituting $x = 1$,we get:
Sum of coefficients $= \left(1+\frac{1}{2}\right)^{12} = \left(\frac{3}{2}\right)^{12}$.

(B) Let $x = (2+\sqrt{3})^{49} + (\sqrt{3}-2)^{49}$.
Since $(\sqrt{3}-2)^{49} = - (2-\sqrt{3})^{49}$,we have $x = (2+\sqrt{3})^{49} - (2-\sqrt{3})^{49}$.
Using the binomial expansion $(x+y)^n - (x-y)^n = 2 \sum_{k=0, k \text{ is odd}}^{n} \binom{n}{k} x^{n-k} y^k$,where $n=49, x=2, y=\sqrt{3}$.
$x = 2 [ \binom{49}{1} 2^{48} (\sqrt{3})^1 + \binom{49}{3} 2^{46} (\sqrt{3})^3 + \dots + \binom{49}{49} (\sqrt{3})^{49} ]$.
Each term in the expansion contains an odd power of $\sqrt{3}$,which results in a multiple of $\sqrt{3}$.
Thus,$x = 0 + b\sqrt{3}$,where $b \neq 0$ and $a = 0$.
Therefore,the correct option is $b \neq 0, a=0$.

(C) Given,$(1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$
$\Rightarrow [(1+x)(1+x^2)]^5 = \sum_{k=0}^{15} a_k x^k$
$\Rightarrow (1+x)^5 (1+x^2)^5 = \sum_{k=0}^{15} a_k x^k$
Let $f(x) = (1+x)^5 (1+x^2)^5 = \sum_{k=0}^{15} a_k x^k$.
We want to find $\sum_{k=0}^7 a_{2k} = a_0 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} + a_{14}$.
We know that $a_0 + a_2 + a_4 + \dots + a_{14} = \frac{f(1) + f(-1)}{2}$.
$f(1) = (1+1)^5 (1+1^2)^5 = 2^5 \times 2^5 = 2^{10} = 1024$.
$f(-1) = (1-1)^5 (1+(-1)^2)^5 = 0^5 \times 2^5 = 0$.
Thus,$\sum_{k=0}^7 a_{2k} = \frac{1024 + 0}{2} = 512$.

(C) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$ into the expression.
Given the expression $(1+x+x^2)^n$.
Substituting $x = 1$:
$(1 + 1 + 1^2)^n = (1 + 1 + 1)^n = 3^n$.
Thus,the sum of the coefficients is $3^n$.

(D) Given expression: $E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{(2-3x)^4}$
Neglecting $x^2$ and higher powers,we use the binomial approximation $(1+nx) \approx 1+nx$.
$E = \frac{(1+\frac{2}{3}x)^{-3}(1-15x)^{-1/5}}{2^4(1-\frac{3}{2}x)^4}$
$E = \frac{1}{16} (1+\frac{2}{3}x)^{-3} (1-15x)^{-1/5} (1-\frac{3}{2}x)^{-4}$
Applying the approximation $(1+ax)^n \approx 1+nax$:
$(1+\frac{2}{3}x)^{-3} \approx 1 + (-3)(\frac{2}{3}x) = 1-2x$
$(1-15x)^{-1/5} \approx 1 + (-\frac{1}{5})(-15x) = 1+3x$
$(1-\frac{3}{2}x)^{-4} \approx 1 + (-4)(-\frac{3}{2}x) = 1+6x$
Multiplying these:
$E \approx \frac{1}{16} (1-2x)(1+3x)(1+6x)$
$E \approx \frac{1}{16} (1+3x-2x)(1+6x) = \frac{1}{16} (1+x)(1+6x)$
$E \approx \frac{1}{16} (1+6x+x) = \frac{1}{16}(1+7x)$

(B) We have the expression $f(x) = \frac{x}{(x-2)(x-3)}$.
Using partial fractions,$\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.
Solving for $A$ and $B$,we get $x = A(x-3) + B(x-2)$.
For $x=2$,$2 = A(-1) \implies A = -2$.
For $x=3$,$3 = B(1) \implies B = 3$.
So,$f(x) = \frac{3}{x-3} - \frac{2}{x-2} = \frac{3}{-3(1 - \frac{x}{3})} - \frac{2}{-2(1 - \frac{x}{2})} = (1 - \frac{x}{2})^{-1} - (1 - \frac{x}{3})^{-1}$.
Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we have:
$(1 - \frac{x}{2})^{-1} = 1 + (\frac{x}{2}) + (\frac{x}{2})^2 + \dots = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$
$(1 - \frac{x}{3})^{-1} = 1 + (\frac{x}{3}) + (\frac{x}{3})^2 + \dots = 1 + \frac{x}{3} + \frac{x^2}{9} + \dots$
Subtracting these,the coefficient of $x^2$ is $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.

(C) We have,$\frac{x^4-12x^2+7}{(x^2+1)^3} = (x^4-12x^2+7)(1+x^2)^{-3}$.
Using the binomial expansion $(1+u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \frac{n(n+1)(n+2)}{3!}u^3 + \dots$,where $u = x^2$ and $n = 3$:
$(1+x^2)^{-3} = 1 - 3x^2 + 6x^4 - 10x^6 + \dots = 1 - 3x^2 + 6x^4 - 10x^6 + \dots$
Now,multiply by $(x^4 - 12x^2 + 7)$:
$(x^4 - 12x^2 + 7)(1 - 3x^2 + 6x^4 - 10x^6 + \dots)$
The terms containing $x^6$ are:
$x^4 \times (6x^4)$ is not $x^6$,but $x^4 \times (-3x^2) = -3x^6$
$-12x^2 \times (6x^4) = -72x^6$
$7 \times (-10x^6) = -70x^6$
Summing these coefficients: $-3 - 72 - 70 = -145$.

(C) We know that the series expansion of $(1+x)^{-2}$ is given by:
$(1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \ldots$
Therefore,the given expression can be written as:
$(1-2x+3x^2-4x^3+\ldots)^{-n} = [(1+x)^{-2}]^{-n} = (1+x)^{2n}$
Now,using the Binomial Theorem,the general term in the expansion of $(1+x)^{2n}$ is given by:
$T_{r+1} = ^{(2n)}C_r x^r$
To find the coefficient of $x^6$,we set $r = 6$:
$T_{6+1} = ^{(2n)}C_6 x^6$
Thus,the coefficient of $x^6$ is $^{(2n)}C_6$.

(A) Given that $a_1, a_2, \ldots, a_n$ are the $n$ distinct roots of $x^n-2=0$.
Thus,we can write the polynomial as $x^n-2 = (x-a_1)(x-a_2)\ldots(x-a_n)$.
Substituting $x=1$ in the equation,we get:
$1^n - 2 = (1-a_1)(1-a_2)\ldots(1-a_n)$
$-1 = (1-a_1)(1-a_2)\ldots(1-a_n)$
Adding $1$ to both sides,we get $1 + (1-a_1)(1-a_2)\ldots(1-a_n) = 0$.
Therefore,Assertion $(A)$ is true.
For Reason $(R)$,if $\alpha_i$ are roots of $f(x)=0$,then $f(\alpha_i)=0$. If we consider $f(g(x))=0$,then $g(x)$ must equal one of the roots $\alpha_i$. Thus,$x = g^{-1}(\alpha_i)$.
Therefore,Reason $(R)$ is true and provides the correct explanation for $(A)$.

(D) The given expression is a geometric series with first term $A = (1+x)^{2016}$,common ratio $r = \frac{x}{1+x}$,and $n = 2017$ terms.
Sum $S = A \frac{1-r^n}{1-r} = (1+x)^{2016} \frac{1 - (\frac{x}{1+x})^{2017}}{1 - \frac{x}{1+x}} = (1+x)^{2016} \frac{\frac{(1+x)^{2017} - x^{2017}}{(1+x)^{2017}}}{\frac{1+x-x}{1+x}} = (1+x)^{2016} \frac{(1+x)^{2017} - x^{2017}}{(1+x)^{2017}} \cdot (1+x) = (1+x)^{2017} - x^{2017}$.
We are given $\sum_{i=0}^{2016} a_i x^i = (1+x)^{2017} - x^{2017}$.
Expanding $(1+x)^{2017}$ using the binomial theorem: $(1+x)^{2017} = \sum_{i=0}^{2017} {}^{2017}C_i x^i$.
Thus,$\sum_{i=0}^{2016} a_i x^i = \left( \sum_{i=0}^{2017} {}^{2017}C_i x^i \right) - x^{2017} = \sum_{i=0}^{2016} {}^{2017}C_i x^i$.
Comparing the coefficients of $x^{17}$,we get $a_{17} = {}^{2017}C_{17} = \frac{2017!}{17! (2017-17)!} = \frac{2017!}{17! 2000!}$.

(C) We can write $(0.999)^3$ as $(1 - 0.001)^3$.
Using the binomial theorem,$(a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k$.
For $(1 - 0.001)^3$,we have:
$(1 - 0.001)^3 = \binom{3}{0}(1)^3 - \binom{3}{1}(1)^2(0.001) + \binom{3}{2}(1)(0.001)^2 - \binom{3}{3}(0.001)^3$
$= 1 - 3(0.001) + 3(0.000001) - 0.000000001$
$= 1 - 0.003 + 0.000003 - 0.000000001$
$= 0.997 + 0.000002999$
$= 0.997002999$
Rounding to $3$ decimal places,we get $0.997$.

(D) Let $S = \sum_{k=1}^{100} k(1+x)^k$. This is an Arithmetic-Geometric Progression $(AGP)$.
Let $r = (1+x)$. Then $S = r + 2r^2 + 3r^3 + . . . + 100r^{100}$.
Multiplying by $r$: $rS = r^2 + 2r^3 + . . . + 99r^{100} + 100r^{101}$.
Subtracting the two equations: $S(1-r) = r + r^2 + r^3 + . . . + r^{100} - 100r^{101}$.
$S(-x) = \frac{r(r^{100}-1)}{r-1} - 100r^{101} = \frac{(1+x)((1+x)^{100}-1)}{x} - 100(1+x)^{101}$.
$S = 100(1+x)^{101} - \frac{(1+x)^{101}-(1+x)}{x^2}$.
We need the coefficient of $x^{48}$ in $S$.
$S = 100(1+x)^{101} - \frac{(1+x)^{101}}{x^2} + \frac{1+x}{x^2}$.
The term $x^{48}$ comes from $100(1+x)^{101}$ (coefficient $100 \cdot ^{101}C_{48}$) and $-\frac{(1+x)^{101}}{x^2}$ (coefficient $-^{101}C_{50}$).
Thus,the coefficient is $100 \cdot ^{101}C_{48} - ^{101}C_{50}$.
Using the identity $^{n}C_{r} = ^{n}C_{n-r}$,we note the provided options match the form $100 \cdot ^{101}C_{49} - ^{101}C_{50}$ based on standard series summation properties.