Binomial Theorem
The coefficient of ${x^{301}}$ in ${(1 + x)^{500}} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$ is :
Let K be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+x)^{99}$. Let $a$ be the middle term in the expansion of ${\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}}$. If ${{{}^{200}{C_{99}}K} \over a} = {{{2^l}m} \over n}$, where m and n are odd numbers, then the ordered pair $(l,\mathrm{n})$ is equal to
If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1 + x)^{10}$, then $\sum\limits_{r = 1}^{10} {{r^3}{{\left( {{{{a_r}} \over {{a_{r - 1}}}}} \right)}^2}} $ is equal to
If ${({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2} = {{\alpha 60!} \over {{{(30!)}^2}}}$ then $\alpha$ is equal to :
The value of $\sum\limits_{r = 0}^{22} {{}^{22}{C_r}{}^{23}{C_r}} $ is
The remainder, when $7^{103}$ is divided by 17, is __________
Explanation:
$ \begin{aligned} & =7 \times(49)^{51} \\\\ & =7 \times(51-2)^{51} \end{aligned} $
Remainder = $7 \times(-2)^{51}$
$ \begin{aligned} & =-7\left(2^3 \cdot(16)^{12}\right) \\\\ & =-56(17-1)^{12} \end{aligned} $
Remainder $=-56 \times(-1)^{12}=-56+68=12$
Let $\alpha$ be the constant term in the binomial expansion of $\left(\sqrt{x}-\frac{6}{x^{\frac{3}{2}}}\right)^{n}, n \leq 15$. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{-n}$ is $\lambda \alpha$, then $\lambda$ is equal to _____________.
Explanation:
The $r$-th term of the binomial expansion of $(a+b)^n$ is given by
$T_{r} = {}^n{C_r}a^{n-r}b^{r}$.
Substitute $a$ and $b$ in this formula, we get:
$T_{r} = {}^n{C_r}(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3/2}}\right)^r = {}^n{C_r}(-6)^r x^{\frac{n-4r}{2}}$.
The constant term in the binomial expansion is obtained when the power of $x$ in the terms equals zero.
This happens when $\frac{n-4r}{2} = 0$, which gives $n = 4r$.
$ \begin{aligned} & { }^n C_{\frac{n}{4}}(-6)^{\frac{n}{4}}=\alpha \\\\ & (-5)^n-{ }^n C_{\frac{n}{4}}(-6)^{n / 4}=649 \\\\ & \text { By observation (625 + 24 = 649) , we get n = 4 } \\\\ & \therefore \alpha=-24 \end{aligned} $
Now, for coefficient of $x^{-4}$
$ \begin{aligned} & \frac{n-4 r}{2}=-4 \\\\ & n=4 r-8 \Rightarrow r=3 \\\\ & \lambda(-24)=(-6)^3 \cdot{ }^4 C_3 \\\\ & \Rightarrow \lambda=36 \end{aligned} $
The mean of the coefficients of $x, x^{2}, \ldots, x^{7}$ in the binomial expansion of $(2+x)^{9}$ is ___________.
Explanation:
$ T_{r+1}={ }^n C_r 2^{n-r} \times x^r $
Coefficient of $x\left(T_1\right)={ }^9 C_1 \times 2^8$
Coefficient of $x^2\left(T_2\right)={ }^9 C_2 \times 2^7$
Coefficient of $x^3\left(T_3\right)={ }^9 C_3 \times 2^6$
              .                    .
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Coefficient of $x^7\left(T_7\right)={ }^9 C_7 \times 2^2$
$ \text { Mean }=\frac{{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2}{7} $
$ \begin{aligned} & { }^9 C_0 \times 2^9+{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots .+{ }^9 C_7 \times 2^2 \\ & =\frac{+{ }^9 C_8 \times 2^1+{ }^9 C_9 \times 2^0-{ }^9 C_0 \times 2^9-{ }^9 C_8 \times 2^1-{ }^9 C_9 \times 2^0}{7} \end{aligned} $
$ =\frac{(1+2)^9-2^9-18-1}{7}=\frac{19152}{7}=2736 $
The number of integral terms in the expansion of $\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$ is equal to ___________.
Explanation:
The term will be integral if $r$ is a multiple of 4 .
$ \begin{gathered} \therefore r=0,4,8,12, \ldots, 680(\text { which is an } \mathrm{AP}) \\\\ 680=0+(n-1) 4 \\\\ n=\frac{680}{4}+1=171 \end{gathered} $
The coefficient of $x^7$ in ${(1 - x + 2{x^3})^{10}}$ is ___________.
Explanation:
So, general term is $\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}(1)^{r_1}(-1)^{r_2} \cdot(2)^{r_3} \cdot(x)^{r_2+r_3}$
Where, $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$
Now, for possibility,
$\begin{array}{ccc}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 7 & 1 & 2 \\ 5 & 4 & 1\end{array}$
Thus, required co-efficient
$ \begin{aligned} & =\frac{10 !}{3 ! 7 !}(-1)^7+\frac{10 !}{5 ! 4 !}(-1)^4(2)+\frac{10 !}{7 ! 2 !}(-1)^1(2)^2 \\\\ & =-120+2520-1440 \\\\ & =2520-1560=960 \end{aligned} $
Let $[t]$ denote the greatest integer $\leq t$. If the constant term in the expansion of $\left(3 x^{2}-\frac{1}{2 x^{5}}\right)^{7}$ is $\alpha$, then $[\alpha]$ is equal to ___________.
Explanation:
$ \mathrm{T}_{r+1}={ }^7 \mathrm{C}_r\left(3 x^2\right)^{7-r}\left(\frac{-1}{2 x^5}\right)^r $
For constant term, power of $x$ should be zero.
$ \begin{aligned} & \text { i.e., } 14-2 r-5 r=0 \\\\ & \Rightarrow 14=7 r \Rightarrow r=2 \end{aligned} $
Now, constant term $=\alpha$
$ \begin{aligned} & \Rightarrow{ }^7 C_2(3)^5\left(\frac{-1}{2}\right)^2=\alpha \\\\ & \Rightarrow 21 \times 243 \times \frac{1}{4}=\alpha \\\\ & \Rightarrow[\alpha]=[1275.75]=1275 \end{aligned} $
The largest natural number $n$ such that $3^{n}$ divides $66 !$ is ___________.
Explanation:
$ \begin{aligned} & {\left[\frac{66}{3}\right]=22} \\\\ & {\left[\frac{66}{3^2}\right]=7} \\\\ & {\left[\frac{66}{3^3}\right]=2} \end{aligned} $
Highest powers of 3 is greater than 66. So, their g.i.f. is always 0.
$\therefore$ Required natural number $=22+7+2=31$
The coefficient of $x^{18}$ in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ is __________.
Explanation:
$\begin{aligned} \therefore 60-7 r =18 \\\\ \Rightarrow 7 r =42 \\\\ \Rightarrow r =6\end{aligned}$
$\therefore$ The coefficient of $x^{18}$
$ ={ }^{15} C_6(-1)^6=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1}=5005 $
Let the sixth term in the binomial expansion of ${\left( {\sqrt {{2^{{{\log }_2}\left( {10 - {3^x}} \right)}}} + \root 5 \of {{2^{(x - 2){{\log }_2}3}}} } \right)^m}$ in the increasing powers of $2^{(x-2) \log _{2} 3}$, be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $x$ is __________.
Explanation:
$ \begin{aligned} \therefore \quad & a={ }^m C_1 \\\\ & a+2 d={ }^m C_2 \\\\ & a+4 d={ }^m C_3 \\\\ \therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\ \Rightarrow & m=7 \text { or } m=2 \\\\ \because & m=2 \text { is not possible } \\\\ \therefore & m=7 \end{aligned} $
$\mathrm{T}_6={ }^{\mathrm{m}} \mathrm{C}_5\left(10-3^{\mathrm{x}}\right)^{\frac{\mathrm{m}-5}{2}} \cdot\left(3^{\mathrm{x}-2}\right)=21$
Putting value of m = 7, we get
$\begin{aligned} & T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21 \\\\ & \Rightarrow \frac{10.3^x-\left(3^x\right)^2}{3^2}=1\end{aligned}$
$ \begin{aligned} & \Rightarrow \left(3^x\right)^2-10 \cdot 3^x+9=0 \\\\ & \Rightarrow 3^x=9,1 \\\\ & \Rightarrow x=0,2 \end{aligned} $
Sum of squares of values of x = 02 + 22 = 4
If the term without $x$ in the expansion of $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22}$ is 7315 , then $|\alpha|$ is equal to ___________.
Explanation:
$ T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r $
For constant term
$ \begin{aligned} & \frac{44-2 r}{3}-3 r=0 \\\\ & \Rightarrow r=4 \end{aligned} $
Now ${ }^{22} \mathrm{C}_4 \alpha^4=7315$
$ \begin{aligned} & \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315 \\\\ & \therefore \alpha^4=1 \\\\ & \therefore |\alpha|=1 \end{aligned} $
The remainder, when $19^{200}+23^{200}$ is divided by 49 , is ___________.
Explanation:
= $(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201}$
Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
$=49 \lambda+470$
$=49(\lambda+9)+29$
$ \therefore $ Remainder $=29$
expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$, is
Explanation:
$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{4 x}{5}\right)^{9-r}\left(\frac{5}{2 x^{2}}\right)^{r} \\\\ 9-3 r & =-6 \\\\ r & =5 \end{aligned} $
Coeff of $x^{-6}={ }^{9} C_{5}\left(\frac{4}{5}\right)^{4}\left(\frac{5}{2}\right)^{5}$ $ =5040 $
Explanation:
$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r}\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^{l}}\right)^{r} \\ & ={ }^{9} C_{r} x^{\frac{45-5 r}{2}-lr} \cdot 2^{r-9} \cdot 4^{r} \cdot(-1)^{r} \end{aligned} $
For constant term, power of x is zero.
So, $\frac{45-5 r}{2}=l r \Rightarrow 2 l r+5 r=45$
Now constant term $=-84$
and ${ }^{9} C_{r} \cdot 2^{3 r-9}(-1)^{r}=-84$
So, $r=3$ and $l=5$
Now for $x^{-15}, \frac{45-5 r}{2}-5 r=-15$
$ \Rightarrow $ $ 45-15 r=-30 $
$ \Rightarrow $ $ r=5 $
$\therefore $ Coefficient $=-{ }^{9} C_{5} 2^{6}=-63.2^{7}$
$\therefore \alpha=7, \beta=-63$
and $|\alpha l-\beta|=|7 \times 5+63|=98$
The remainder on dividing $5^{99}$ by 11 is ____________.
Explanation:
$=625\left[5^{5}\right]^{19}$
$=625[3125]^{19}$
$=625[3124+1]^{19}$
$=625[11 \mathrm{k} \times 19+1]$
$=625 \times 11 \mathrm{k} \times 19+625$
$=11 \mathrm{k}_{1}+616+9$
$=11\left(\mathrm{k}_{2}\right)+9$
Remainder $=9$
Let $\alpha>0$, be the smallest number such that the expansion of $\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}$ has a term $\beta x^{-\alpha}, \beta \in \mathbb{N}$. Then $\alpha$ is equal to ___________.
Explanation:
$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
$\frac{60-11 \mathrm{r}}{3}<0 $
$\Rightarrow 11 \mathrm{r}>60 $
$\Rightarrow \mathrm{r}>\frac{60}{11} $
$\Rightarrow \mathrm{r}=6$
$\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}$
We have also observed $\beta={ }^{30} \mathrm{C}_{6}(2)^{6}$ is a natural number.
$\therefore \alpha=2$
Explanation:
Given ${x^{{1 \over {50}}}} = 12 \Rightarrow x = {12^{50}}$
${y^{{1 \over {50}}}} = 18 \Rightarrow y = {18^{50}}$
$12\equiv13$ (Mod 25)
$12^2\equiv19$ (Mod 25)
$12^3\equiv-3$ (Mod 25)
$12^9\equiv-2$ (Mod 25)
$12^{10}\equiv-1$ (Mod 25)
$12^{50}\equiv-1$ (Mod 25) ..... (i)
Now
$18\equiv7$ (Mod 25)
$18^2\equiv-1$ (Mod 25)
$18^{-50}\equiv-1$ (Mod 25) ..... (ii)
$\therefore$ $12^{50}+18^{50}\equiv-2$ (Mod 25)
$\equiv23$ (Mod 25)
$\therefore$ Answer = 23
Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.
Explanation:
$ \begin{aligned} & \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\ & \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\ & \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{r}+1}}=\frac{5}{8} \\\\ & \Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \\\\ & \Rightarrow 4 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2) \end{aligned} $
From (1) and (2)
$ \Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8 $
$(1) \Rightarrow r=4$
so, coefficient of middle term is
$ { }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120 $
If the co-efficient of $x^9$ in ${\left( {\alpha {x^3} + {1 \over {\beta x}}} \right)^{11}}$ and the co-efficient of $x^{-9}$ in ${\left( {\alpha x - {1 \over {\beta {x^3}}}} \right)^{11}}$ are equal, then $(\alpha\beta)^2$ is equal to ___________.
Explanation:
$\because$ Both are equal
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
$\Rightarrow \frac{1}{\beta}=-\alpha$
$\Rightarrow \alpha \beta=-1$
$\Rightarrow(\alpha \beta)^{2}=1$
The remainder when (2023)$^{2023}$ is divided by 35 is __________.
Explanation:
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7
The constant term in the expansion of ${\left( {2x + {1 \over {{x^7}}} + 3{x^2}} \right)^5}$ is ___________.
Explanation:
$ \begin{aligned} & \left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(2 x^{8}+1+3 x^{9}\right)^{5} \\\\ & \frac{1}{x^{35}}\left(1+x^{8}(3 x+2)\right)^{5} \end{aligned} $
Term independent of $x=$ coefficient of $x^{35}$ in
$ \begin{aligned} & ^{5} C_{4}\left(x^{8}(3 x+2)\right)^{4} \\\\ = & { }^{5} C_{4} \text { coefficient of } x^{3} \text { in }(2+3 x)^{4} \\\\ = & { }^{5} C_{4} \times{ }^{4} C_{3}(2)^{1}(3)^{3} \\\\ = & 5 \times 4 \times 2 \times 27 \\\\ = & 1080 \end{aligned} $
Let the sum of the coefficients of the first three terms in the expansion of ${\left( {x - {3 \over {{x^2}}}} \right)^n},x \ne 0.~n \in \mathbb{N}$, be 376. Then the coefficient of $x^4$ is __________.
Explanation:
$ \begin{aligned} & 3 n^{2}-5 n-250=0 \\\\ & n=10, \frac{-25}{3} \text { (Rejected) } \\\\ & T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\ & ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\ & ={ }^{10} C_{r} x^{10-3 r}(-3)^{r} \end{aligned} $
Here $r=2$
$ \begin{aligned} \text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\ & =45 \times 9 \\\\ & =405 \end{aligned} $
Suppose $\sum\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \times \alpha \times {2^{2022}}} $. Then the value of $\alpha$ is ___________
Explanation:
Concept :
(1) ${}^n{C_r} = {n \over r}\,.\,{}^{n - 1}{C_{r - 1}}$
Given,
$\sum\limits_{r = 0}^{2023} {{r^2}\,.\,{}^{2023}{C_r}} $
$ = \sum\limits_{r = 0}^{2023} {{r^2}\,.\,{{2023} \over r}\,.{}^{2022}{C_{r - 1}}} $
$ = 2023\sum\limits_{r = 0}^{2023} {{r}\,.\,{}^{2022}{C_{r - 1}}} $
$ = 2023\sum\limits_{r = 0}^{2023} {[(r - 1) + 1]\,.\,{}^{2022}{C_{r - 1}}} $
$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{}^{2022}{C_{r - 1}} + \sum\limits_{r = 0}^{2023} {{}^{2022}{C_{r - 1}}} } } \right]$
$ = 2023\left[ {\sum\limits_{r = 0}^{2023} {(r - 1)\,.\,{{2022} \over {(r - 1)}}\,.\,{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$
$ = 2023\left[ {2022\sum\limits_{r = 0}^{2023} {{}^{2021}{C_{r - 2}} + {2^{2022}}} } \right]$
$ = 2023\left[ {2022\,.\,{2^{2021}} + {2^{2022}}} \right]$
$ = 2023\,.\,{2^{2021}}\left[ {2022 + 2} \right]$
$ = 2023\,.\,{2^{2021}}\,.\,2024$
$ = 2023\,.\,{{{2^{2022}}} \over {{2}}}\,.\,2024$
$ = 2023\,.\,{2^{2022}}\,.\,{1012}$
$\therefore$ $\alpha = {1012}$
The term independent of $x$ in the expansion of $\left(1-3 x+2 x^3\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$ is
$7 / 18$
$5 / 18$
$19 / 54$
$17 / 54$
If $\sum_{r=0}^{20}{ }^{20+r} C_r=\frac{p}{q}{ }^{40} C_{20}$ and GCD of $(p, q)=1$, then $p^2-q^2=$
1302
1220
1240
1364
If $x=\frac{2 \cdot 5}{2!3}+\frac{2 \cdot 5 \cdot 7}{3!3^2}+\frac{2 \cdot 5 \cdot 7 \cdot 9}{4!3^3}+\ldots$, then $x^2+8 x+8=$
108
54
100
144
If the coefficient of $x^4$ in the expansion of $\frac{x}{(x-1)^2(x-2)}$ is $\frac{m}{n}$ and $|m|,|n|$ are coprimes, then $\sqrt{|m+n|}=$
9
$\sqrt{65}$
7
$6 \sqrt{2}$
If $(-c, c)$ is the set of all values of $x$ for which the expansion of $(7-5 x)^{\frac{-2}{3}}$ is valid, then $5 c+7=$
0
12
41
14
If $n$ is a positive integer and $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$, then $f(2023)=$
-2021
2022
2023
-2023
If $y=\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots$ to $\infty$, then
$y^2-2 y+5=0$
$y^2+2 y-7=0$
$y^2-3 y+4=0$
$y^2+4 y-6=0$
The numerically greatest term in the binomial expansion of $(2 x-3 y)^5$, when $x=\frac{3}{2}$ and $y=\frac{2}{3}$ is
360
1080
720
2160
If $\frac{2 x^3+3 x^2+3 x+5}{\left(x^2+1\right)\left(x^2+2\right)}$ is expanded in terms of the powers of $x$, then the coefficient of $x^5$ is
0
$\frac{-5}{4}$
$\frac{17}{8}$
$\frac{9}{8}$
In the expansion of $(x-2 y+3 z)^5$, if the total number of terms is $p$ and the coefficient of $x^2 y z^2$ is $q$, then $\frac{q}{p}=$
60
$-\frac{180}{7}$
72
$-\frac{1080}{7}$
Let $C_0, C_1, C_2, \ldots, C_n$ be the binomial coefficients in the expansion of $(1+x)^n$. If $S_{n+1}=5 \cdot C_0+8 \cdot C_1+11 \cdot C_2+\ldots(n+1)$ terms, then $S_{11}=$
18944
17920
20480
40960
If $|x|$ is so small that $x^3$ and higher powers of $x$ can be neglected, then an approximate value of $\frac{1}{\sqrt{4-x}(2+x)^3}$ is
$\frac{1}{16}\left(1+\frac{13 x}{8}+\frac{219}{128} x^2\right)$
$\frac{1}{8}\left(1+\frac{11 x}{8}-\frac{165}{128} x^2\right)$
$\frac{1}{32}\left(1-\frac{11 x}{8}+\frac{219}{128} x^2\right)$
$\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right)$
The number of integral terms in the expansion of $(\sqrt{3}+\sqrt[8]{5})^{256}$ is
The expansion of $\left(1+x+x^2\right)^{-3 / 2}$ in powers of $x$ is valid, if
$-\frac{1}{2}-\frac{\sqrt{5}}{2} < x < 1$
If $(1+x)^n=c_0+c_1 x+c_2 x^2+\ldots \ldots+c_n x^n$ for $n \in N$, then $c_0+\frac{c_1}{2}+\frac{c_2}{3}+\ldots \ldots+\frac{c_n}{n+1}=$
The sum of the coefficients of all odd degree terms in the expansion of $\left(x+\sqrt{x^3-1}\right)^5 +\left(x-\sqrt{x^3-1}\right)^5, x>1$ is
$\sum_\limits{\substack{i, j=0 \\ i \neq j}}^n{ }^n C_i{ }^n C_j$ is equal to
$\sum\limits_{r=1}^{20}\left(r^{2}+1\right)(r !)$ is equal to
The remainder when $7^{2022}+3^{2022}$ is divided by 5 is :
The remainder when $(2021)^{2022}+(2022)^{2021}$ is divided by 7 is