2 Marks Questions

Question 12 Marks

If $C _0, C _1, C _2, \ldots \ldots . C _n$ are the binomial coefficients in the expansion of $(1+x)^n$, then prove that :
$C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\ldots \ldots \ldots+\frac{(-1)^n \cdot C_n}{n+1}=\frac{1}{n+1}$

Answer

L.H.S. $= C _0-\frac{ C _1}{2}+\frac{ C _2}{3}-\frac{ C _3}{4}+\ldots \ldots .+\frac{(-1)^n \cdot C _n}{n+1}$
$=1-\frac{1}{2} n+\frac{1}{3} \cdot \frac{n(n-1)}{2!}-\frac{1}{4} \cdot \frac{n(n-1)(n-2)}{3!}+\ldots \ldots+\frac{(-1)^n}{n+1} \cdot 1\ $
$=\frac{1}{n+1}$
$ \left[(n+1)-\frac{(n+1) n}{2!}+\frac{(n+1) n(n-1)}{3!} \ldots .+(-1)^n \cdot 1\right] $
[Multiplying and dividing by $(n+1)$]
Now, putting $m=n+1$
$=\frac{1}{m}\left[m-\frac{m(m-1)}{2!}+\frac{m(m-1)(m-2)}{3!} \ldots . .+(-1)^{m-1}\right]$
$=\frac{1}{m}\left[{ }^m C _1-{ }^m C _2+{ }^m C _3-\ldots \ldots . .+(-1)^{m-1} \cdot{ }^m C _m\right]$
$=\frac{1}{m}\left[1-\left\{1-{ }^m C _1+{ }^m C _2-{ }^m C _3+\ldots .+(-1)^{m-1} \cdot{ }^m C _m\right]\right.$
[Adding and subtracting 1]
$=\frac{1}{m}\left[1-(1-1)^m\right]=\frac{1}{m}[1-0]=\frac{1}{m}$
Putting the value of $m$
$ =\frac{1}{n+1}=\text { R.H.S. } $

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Question 22 Marks

If $C_0, C_1, C_2, \ldots \ldots \ldots . C_n$ are the binomial coefficients in the expansion of $(1+x)^n$, then prove that :
$2 C_0+\frac{2^2 C_1}{2}+\frac{2^3 C_2}{3}+\ldots \ldots +\frac{2^{n+1} \cdot C_n}{n+1}=\frac{3^{n+1}-1}{n+1}$

Answer

$\begin{array}{l}\text { L.H.S. }=2 C _0+\frac{2^2 C _1}{2}+\frac{2^3 C _2}{3}+\ldots \ldots+\frac{2^{n+1} \cdot C _n}{n+1} \\ =2 \cdot 1+\frac{2^2}{2} \cdot n+\frac{2^3}{3} \cdot \frac{n(n-1)}{2!}+\ldots .+\frac{2^{n+1}}{n+1} \cdot 1 \\ =\frac{1}{(n+1)}\end{array}$
$\left[(n+1) \cdot 2+\frac{(n+1) n}{2!} \cdot 2^2+\frac{(n+1) n(n-1)}{3!} \cdot 2^3+\ldots \cdot 2^{n+1}\right]$
Now, putting $n+1=m$
$=\frac{1}{m}\left[m \cdot 2+\frac{m(m-1)}{2!} \cdot 2^2+\frac{m(m-1)(m-2)}{3!} \cdot 2^3+\ldots 2^m\right]$
Adding and subtracting ${ }^m C _0$
$\begin{array}{r}=\frac{1}{m}\left[{ }^m C_0+{ }^m C_1 \cdot 2+{ }^m C_2 \cdot 2^2+{ }^m C_3 \cdot 2^3\right. \left.+\ldots \ldots \cdot+{ }^m C_m \cdot 2^m-{ }^m C_0\right] \end{array}$
$\begin{array}{l}=\frac{1}{m}\left[(1+2)^m-1\right]=\frac{1}{m}\left(3^m-1\right) \\ {\left[\because{ }^m C _0=1\right]}\end{array}$
$=\frac{3^{n+1}-1}{n+1}=$ R.H.S. (Putting the value of $m$ )

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Question 32 Marks

If the expansion of $\left(1+x-2 x^2\right)^6$ is $1+a_1 x+$ $a_2 x^2+a_3 x^3+$ $\ldots\ldots\ldots$ $+a_{12} x^{12}$, then prove that $a_2$ $+a_4+a_6+\ldots \ldots \ldots .+a_{12}=31$.

Answer

In the given expansion putting $x=1$ and $x=-1$.
$\begin{array}{l}\left(1+1-2 \times 1^2\right)^6=1+a_1+a_2+a_3+a_4 +\ldots \ldots \ldots +a_{12}\end{array}$
$\begin{array}{l}0=1+a_1+a_2+a_3+a_4 +\ldots \ldots \ldots +a_{12} \ldots . . \text { (1) }\end{array}$
$\begin{aligned}(1-1-2 \times 1)^6=1-a_1 & +a_2-a_3+a_4 - \ldots \ldots \ldots + a_{12} \end{aligned}$
$\begin{array}{l}64=1-a_1+a_2-a_3+a_4 - \ldots \ldots \ldots +a_{12} \ldots . . \text { (2) }\end{array}$
Adding equations (1) and (2),
$\begin{array}{rlrl}0+64 & =2+2 a_2+2 a_4+2 a_6+2 a_8+2 a_{10}+2 a_{12} \\64 & =2\left(1+a_2+a_4+a_6+a_8+a_{10}+a_{12}\right) \\& \Rightarrow \quad \frac{64}{2} =1+a_2+a_4+a_6+a_8+a_{10}+a_{12} \\& \Rightarrow 32 =1+a_2+a_4+a_6+a_8+a_{10}+a_{12} \\& \Rightarrow31 =a_2+a_4+a_6+a_8+a_{10}+a_{12} \quad \text { Hence proved. }\end{array}$

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Question 42 Marks

If $\left(1-x+x^2\right)^4=1+p_1 x+p_2 x^2+\ldots \ldots+p_8 x^8$, then prove that :
$p_2+p_4+p_6+p_8=40$

Answer

Given :
$\left(1-x+x^2\right)^4=1+p_1 x+p_2 x^2+p_3 x^2+\ldots .+p_8 x^8 \ldots\ldots (1)$
Putting $x=1$,
$(1-1+1)^4=1+p_1+p_2+p_3+p_4+\ldots .+p_8$
or $1=1+p_1+p_2+p_3+p_4+\ldots . .+p_8\ldots\ldots (2)$
Again putting $x=-1$ in equation (1),
$\begin{aligned}{\left[1-(-1)+(-1)^2\right]^4=1+} & p_1(-1)+p_2(-1)^2+p_3(-1)^3 +p_4(-1)^4+\ldots \ldots . .+p_8(-1)^8\end{aligned}$
or $3^4=1-p_1+p_2-p_3+p_4-\ldots \ldots .+p_8$
Adding equations (2) and (3),
$1+3^4=2\left(1+p_2+p_4+p_6+p_8\right)$
or $\quad 1+p_2+p_4+p_6+p_8=\frac{1+81}{2}=41$
or $\quad p_2+p_4+p_6+p_8=40$

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Question 52 Marks

Solve $(1+\sqrt{3})^4+(1-\sqrt{3})^4$ by using binomial theorem.

Answer

By binomial theorem :
$ \begin{aligned} (1+\sqrt{3})^4=1+{ }^4 C_1(\sqrt{3})^1+{ }^4 C_2 & (\sqrt{3})^2+{ }^4 C_3(\sqrt{3})^3 +{ }^4 C_4(\sqrt{3})^4 \ldots\ldots (1) \end{aligned} $
$\begin{array}{r}\text {and}\quad (1-\sqrt{3})^4=1-{ }^4 C _1(\sqrt{3})^1+{ }^4 C _2(\sqrt{3})^2-{ }^4 C _3 (\sqrt{3})^3+{ }^4 C _4(\sqrt{3})^4\ldots\ldots (2)\end{array}$
Adding equations (1) and (2),
$\begin{aligned}(1+\sqrt{3})^4+(1-\sqrt{3})^4 & =2\left[1+{ }^4 C _2(\sqrt{3})^2+{ }^4 C _4(\sqrt{3})^4\right] \\ & =2[1+6 \times 3+1 \times 9] \\ & =2(1+18+9)=2 \times 28 \\ & =56 \end{aligned}$

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