For two positive integers $m, n$, if in the expansion of $(1+x)^m(1-x)^n$ the coefficients of $x$ and $x^2$ are 3 and -6 respectively, then what is the value of $m$ ?

(1) 6

(2) 9

(3) 12

(4) 24

$(1+x)^m (1-x)^n =\underset{k=0} {\overset{m}\sum} . { }^m C_k .x^k \underset {k=0}{\overset {n}\sum}. { }^n C_r(-1)^k.. x^k$

coefficient of x $\rightarrow{ }^m C_0\left(-{ }^n C_1\right)+{ }^m C_1.{ }^n C_0$

-n+m=3

co-efficient of $x^2 \rightarrow{ }^m C_0 .{ }^nC_2 + { }^m C_1.(-{ }^n C_1) + { }^m C_2 . { }^n C_0$

$\frac{n(n-1)}{2} - m n + \frac{m(m-1)}{2}=-6$

$n^2-n + m^2 - m - 2 m n=-12$

$\Rightarrow(m-n)^2-(m+n)=-12$

9-(m+n)=-12

$\Rightarrow$ m+n=21

m - n = 3

$\therefore 2m = 24$

$\Rightarrow$ m = 12

The coefficient of $t^{24}$ in the expansion of $(1+t^2)^{12}(1+t^{12})(1+t^{24})$ is

(1) ${ }^{12} C_6$

(2) ${ }^{12} C_6 + 1$

(3) ${ }^{12} C_6 + 2$

(4) ${ }^{12} C_6 + 3$

$(1+t^2)^{12}.(1+t^{12}).(1+t^{24})$

$=\underset{k=0} {\overset {12}\sum} .{ }^{12} C_k .t^{24-2 k} (1+t^{12})(1+t^{24})$

$=\underset {k=0}{\overset {12}\sum} .{ }^{12} C_k (t^{24-2 k}+t^{36-2 k}+t^{48-2 k}+t^{60-2 k})$

${ }^{12} C_0 + { }^{12} C_6 + { }^{12} C_{12}$

$= 1 + { }^{12} C_6 + 1 = 2 + { }^{12} C_6$

The coefficient of $x^{11}$ in the expansion $(1+x^2)^4 (1+x^3)^7(1+X^4)^{12}$ is

(1) 1051

(2) 1106

(3) 1113

(4) 1120

$(1+x^2)^4 = \underset {k=0} {\overset {4}\sum} { }^4 C_k .x^{2k}$ , A = {0,2,4,6,8}

$(1+x^3)^7 = \underset {r=0} {\overset {7}\sum} { }^7 C_r .x^{3r}$ , B = {0,3,6,9}

$(1+x^4)^{12} = \underset {s=0} {\overset {12}\sum} { }^{12} C_s .x^{4s}$ , C = {0,4,8}

Coefficient of $x^{11}$ in the expansion is

${ }^4 C_0 { }^7 C_1 { }^{12} C_2 + { }^4 C_1 { }^7 C_3 { }^{12} C_0 + { }^4 C_2 { }^7 C_1 { }^{12} C_1 + { }^{4} C_4 { }^7 C_1 { }^{12} C_0 = 462 + 140 + 504 + 7 = 1113$

If for some integers $n, r$ such that $0 \leq r

( 1)($-\infty,-2]$

(2) $[2, \infty)$

(3) $[-\sqrt{3}, \sqrt{3}]$

(4) $(\sqrt{3}, 2]$

${ }^{n-1} C_r=(k^2-3) { }^n C_{r+1}$

$\Rightarrow \frac{(n-1) !}{r !(n-1-r) !}=(k^2-3) \frac{n !}{(r+1) !(n-k-1) !}$

$\Rightarrow \frac{k+1}{n}=k^2-3, \quad \frac{1}{n} \leq \frac{k+1}{n} \leq 1$

$0 < \frac{1}{n} \leq k^2 - 3 \leq 1$

$k^2 \leq 4 , k^2 \geq \frac{1}{n} +3 , k^2 > 3$

$k^2 \leq 4 \rightarrow -2 \leq k \leq 2$

$k^2 > 3 \rightarrow k^2 > \sqrt 3 \quad \text{or}\quad k < - \sqrt 3$

$[-2,-\sqrt 3] \cup [\sqrt3,2]$

The remainder of $8^{2020} - 62^{2021}$ when divided by 9 is

(1) 2

(2) 7

(3) 8

(4) 0

$8 = 9-1$

$62 = 63 - 1 \quad [ 63 = 7 \times 9]$

= $8^{2020} - 62^{2021}$

= $(9-1)^{2020} - (63-1)^{2021}$

= $\underset {k=0}{\overset {2020}\sum} { }^{2020}C_k. 9^{2020 - k} (-)^k - \underset{k=0} {\overset {2021} \sum}. { }^{2021} C_r. 63^{2021-r} (-1)^r$

= ${ }^{2020} C_{2020} - { }^{2021} C_{2021} (-1)$

= 2

For a,b$\neq$0, the sum of the $5^{th}$ and the $6^{th}$ terms in the expansion of $(a-b)^n$ is zero for an integer $n\geq5$. Then the ratio a/b is.

(1) $\frac {n-5}{6}$

(2) $\frac {n-4}{5}$

(3) $\frac {5}{n-4}$

(4) $\frac {6}{n-5}$

$\begin{aligned}& (a-b)^n=\sum_{k=0}^n{ }^n C_k a^{n-k}(-1)^k b^k \\ & T_5={ }^n C_4 a^{n-4} b^4 \\ & T_6=-{ }^n C_5 a^{n-5} b^5\end{aligned}$

${ }^n C_4 a^{n-4} b^4-{ }^n C_5 a^{n-5} b^5=0$

$\Rightarrow{ }^n C_4 a^{n-5} b^4 {a-\frac{(n-4)b}{5}} =0$

$\text { As } a, b \neq 0,$

$a-\frac{(n-4) b}{5}=0$

$\Rightarrow \frac{a}{b}=\frac{n-4}{5}$

For a positive integer $n$, the coefficients of three consecutive terms in the expansion of $(1+x)^{n+5}$ are having ratio $5: 10: 14$. Then what is the value of $n$ ?

$(1+x)^{n+5}=\sum_{t=0}^{n+5} n+5 C_t x^t$

$ { }^{n+5} C_{r-1},{ }^{n+5} C_r,{ }^{n+5} C_{r+1} \text { for some } 0

$\biggr[{ }^{n+5} C_{r-1} = 5k \biggr], \ \biggr[{ }^{n+5} C_r = 10k \biggr], \ \biggr[{ }^{n+5} C_{r+1} = 14k \biggr]$

for some $k$

$\frac{(n+5) !}{(r-1) !(n+6-r) !}=5 k$

$\frac{(n+5) !}{r !(n+5-r) !}=10 k$

$\frac{(n+5) !}{(r+1) !(n+4-r) !}=14 k$

$\begin{aligned} & \frac{r}{n-r+6}=\frac{5}{10} \\ & \Rightarrow 2 r=n-r+6 \\ & \Rightarrow 3 r-n=6 \\ & \frac{r+1}{n-r+5}=\frac{10}{14} \\ & \Rightarrow 7 r+7=5 n-5 r+25 \\ & \Rightarrow 12 r-5 n=18 \\ & \end{aligned}$
$12 r-4 n=24$

$12 r-5 n= 18$

$n=6$

Let m be the smallest positive integer such that the coefficient of $x^2$ in the $\operatorname{sum}(1+x)^2+(1+x)^3+\cdots+(1+x)^{49}+(1+m x)^{50}$ is ${ }^{51} C_3(3 n+1)$. Then what is the value of $n$ ?

Solution:

$(1+x)^2+(1+x)^3+\cdots+(1+x)^{49}$

$=(1+x)^2 \{1+(1+x)+\cdots+(1+x)^{47}\}$

$=(1+x)^2 \frac{(1+x)^{48}-1}{x}$

$=\frac{(1+x)^{50}-(1+x)^2}{x}$

$\frac{(1+x)^{50}-(1+x)^2}{x}+{(1+m x)^{50}}$

co-efficient of $x^2$ in the given sum $={}^{50}C_3+m^2 \quad {}^{50}C_2$

${}^{50} C_3+m^{2} \quad {}^{50} C_2={ }^{51}C_3(3 n+1)$

$\Rightarrow(3 n+1)=\frac{{}^{50} C_3}{{}^{51} C_3}+m^2 \frac{{}^{50} C_2}{{}^{51} C_3}$

$\begin{aligned} & 3 n+1=\frac{48}{51}+m^2 \frac{3}{51} \\ & \Rightarrow 153 n+51 =48+3 m^2 \\ & \Rightarrow \quad 3 m^2-3=153 n \\ & m^2-1=51 n \\ & m=1, n=0 \ & \end{aligned}$

For integers $m, n$ with $n \geq m$, show that

${ }^n C_m+{ }^{n-1} C_m+{ }^{n-2} C_m+\cdots+{ }^m C_m={ }^{n+1} C_{m+1}.$

Hence prove that

${}^nC_m+2 {}^{n-1}C_m+3 {}^{n-2}C_m+\cdots+(n-m+1) \quad{}^mC_m={}^{n+2}C_{m+2}$

${ }^m C_m+{ }^{m+1} C_m+{ }^{m+2} C_m+\cdots+{ }^{n-1} C_m+{ }^n C_m$

$[{ }^m C_m+{ }^{m+1} C_m = { }^{m+1} C_{m+1}]$

$={}^{m+2} C_{m+1}+{ }^{m+2} C_m+ \ldots+{ }^{n-1} C_m+{ }^n C_m$

$[{}^{m+2} C_{m+1}+{ }^{m+2} C_m = {}^{m+3}C_{m+1}]$

${ }^{n-1}{C_{m+1}}{ }^{n-1} C_m+{ }^n C_m$

$={ }^{n} C_{m+1}+{ }^n C_m$

$={ }^{n+1} C_{m+1}$

$\begin{aligned}& { }^n C_m+2^{n-1} C_m+3^{n-2} C_m+\cdots+(n-m+1)^m C_m \\& =({}^n C_m+{ }^{n-1} C_m+{ }^{n-2} C_m+\cdots+{ }^m C_m) \\& +\left({ }^{n-1} C_m+{ }^{n-2} C_m+\cdots+{ }^m C_m\right) \\& +\cdots+\left({ }^{m+1} C_m+m C_m\right)+{ }^m C_m \\& ={ }^{n+1} C_{m+1}+{ }^n C_{m+1}+\cdots+{ }^{m+2} C_{m+1}+{ }^{m+1} C_{m+1} \\& ={ }^{n+2} C_{m+2} \text {. } \\&\end{aligned}$

Statement $1: \sum_{r=0}^n(r+1)^n C_r=(n+2) 2^{n-1}$.

Statement 2 : $\sum_{r=0}^n(r+1)^n C_r x^r=(1+x)^n+n x(1+x)^{n-1}$.

• Statement 1 is false, Statement 2 is true.
• Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.
• Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
• Statement 1 is true, Statement 2 is false.

Put $x=1$ in statement 2

$\text { LHS }=\sum_{r=0}^n(r+1)^n C_r$

$\text { RHS }=2^{n}+n 2^{n-1}=2^{n-1}(2+n)$

$(1+x)^n=\sum_{r=0}^n{ }^n C_r x^r$

$x(1+x)^n=\sum_{r=0}^n{ }^n C_r x^{r+1}$

$(1+x)^n+n x(1+x)^{n-1}$

$=\sum_{r=0}^n(r+1)^n C_r x^r$

statement 2 is true.

Let $S_1=\sum_{j=1}^{10} j(j-1){ }^{10} C_j, S_2=\sum_{j=1}^{10} j^{10} C_j$ and $S_3=\sum_{j=1}^{10} j^2{ }^{10} C_j$

Statement 1 : $S_3=55 \times 2^9$

Statement 2 : $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$.

• Statement 1 is false, Statement 2 is true.
• Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.
• Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
• Statement 1 is true, Statement 2 is false.

$S_1+S_2=S_3$

$(1+x)^{10} =\sum_{j=0}^{10}{ }^{10} C_j x^j \\ 10(1+x)^9 =\sum_{j=0}^{10}{ }^{10} C_j j x^{j-1}$
$\text { Put } x =1$
$\text { LHS } =10 \cdot 2^9$
$\text { RHS } =\sum_{j=1}^{10} j^{10} C_j=S_2 \\ S_2 =10 \cdot 2^9$

$10 \cdot 9(1+x)^8=\sum_{j=0}^{10} j(j-1)^{10} C_j x^{j-2}$

Put $x=1$

LHS $=90 \cdot 2^8=2^9 \cdot 45$

$RHS =\sum_{j=2}^{10} j(j-1)^{10} C_j=S_1$

$S_1=2^9 \cdot 45$

$S_3=S_1+S_2=2^9 \cdot 45+2^9 \cdot 10=2^9 \cdot 55$