Binomial Theorem Test - 56

$$ \displaystyle T_{r+1} = ^{1024}C_r\times 5^{\tfrac r2 }\times2^{\tfrac {(1024-r)}8}$$
For $$T_{r+1}$$ to be integral, the powers of $$5$$ and $$2$$ must be integers.
Hence,
$$\displaystyle\frac {r}{2} = a $$ and $$\displaystyle\frac{1024-r}{8} = b $$, where $$a$$ and $$b$$ are integers. 
Hence, $$r$$ must be a factor of $$8$$. 
$$r$$ can assume values from $$0$$ to $$1024$$ which are factors of $$8$$.
 Hence, the total number of values $$r$$ can acquire is $$1 + \displaystyle\frac{1024}{8} = 129$$.

$$(1+x)^{2n+2}$$.
The middle term will be $$(\dfrac {N}{2}+1)$$ th term.
$$=\dfrac {2n+2}{2}+1$$ th term
$$=(n+2)^{th}$$ term
Hence coefficient of the middle term will be
$$\:^{2n+2}C_{n+1}$$
$$=p$$
For 
$$(1+x)^{2n+1}$$.
The middle terms will be $$(\dfrac {N+1}{2}+1)$$ and  $$(\dfrac {N+1}{2})$$ terms
$$=\dfrac {2n+2}{2}+1$$ th term and $$\dfrac {2n+2}{2}^{th}$$ term.
$$=(n+2)^{th}$$ term and $$(n+1)^{th}$$ term.
Hence coefficient of the middle terms will be
$$\:^{2n+1}C_{n+1}=q$$ and $$\:^{2n+1}C_{n}=r$$
By the properties of binomial coefficients.
$$\:^{2n+1}C_{n+1}+\:^{2n+1}C_{n}$$
$$=\:^{2n+2}C_{n+1}$$
Hence $$q+r=p$$

No. of elements in $$b.c.d...  = n$$
Choose $$a^k$$, where k$$ \in {0,1,2, ..m}$$ at a time =$$(m+1) $$
for every $$k$$, no. possible factors from n elements = $$ ^nC_0+^nC_1+^nC_2+ . . . . . + ^nC_n$$
$$ = 2^n$$
Total factors $$ =$$ No. of possible powers of $$a\times$$ every $$k$$, no. possible factors from $$n$$ elements 
$$= (m+1)2^n$$
If factor $$1$$ is excluded then
$$(m+1)2^n-1$$

$$(x^2-2+\dfrac{1}{x^2})^{n}$$
$$=((x-\dfrac{1}{x})^2)^{n}$$
$$=(x-\dfrac{1}{x})^{2n}$$
$$T_{r+1}=\:{10}C_{r}x^{2n-2r}$$
For term independent of $$x$$
$$2n-2r=0$$
$$n=r$$
Hence there will be $$1$$ term independent of $$x$$.
Since the total number of terms are $$2n+1$$.
Hence the total number of term dependent on $$x$$ will be
Total number of terms-(total number of terms independent of $$x$$).
$$=2n+1-1$$
$$=2n$$

$$\dfrac { { c }_{ 0 } }{ x } -\dfrac { { c }_{ 1 } }{ x+1 } +\dfrac { { c }_{ 2 } }{ x+2 } -............\quad \quad +{ \left( -1 \right)  }^{ n }\dfrac { { c }_{ n } }{ x+n } $$

$$c_{0},c_{1},c_{2}\cdots c_{n}$$ are binomial coefficients of $$\left ( 1+x \right )^{n}$$
i.e $$\left ( 1+x \right )^{n} = c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\dots $$
put $$x=i$$
$$\Rightarrow \left ( 1+i \right )^{n} = c_{0}+c_{1}i-c_{2}-c_{3}i+\dots $$
$$\Rightarrow \displaystyle 2^{\frac{n}{2}}c$$ is $${\displaystyle \frac{n\pi}{4}} = c_{0}+c_{1}i-c_{2}-c_{3}i+\dots $$
Comparing imaginary parts gives
$$\displaystyle 2^{\frac{n}{2}}\sin {\dfrac{n\pi}{4}} = c_{1}-c_{3}+c_{5}+\dots $$ ------(1)
$$2^{n-1} = c_{1}+c_{3}+c_{5}+\dots $$  ------(2)       ($$\because\quad c_{0}+c_{1}+c_{2}+\dots = 2^{n}$$)
$$\therefore$$ (1)+(2) gives
$$c_{1}+c_{5}+c_{9}+\dots = \displaystyle\frac{1}{2}\left(2^{n-1}+\displaystyle 2^{\frac{n}{2}}\sin {\dfrac{n\pi}{4}}\right)$$

$$^{45}C_{8}$$+$$\sum _{ k=1 }^{ 7 }{^{ 52-k} C_{ 7 }} $$+$$\sum _{ i=1 }^{ 5 }{^{ 57-i} C_{ 50-i }} $$
$$\Rightarrow $$$$^{45}C_{8}$$+$$^{45}C_{7}$$+$$^{46}C_{7}$$+......+$$^{51}C_{7}$$+$$\sum _{ i=1 }^{ 5 }{^{ 57-i} C_{ 57-i-(50-i) }} $$
$$\Rightarrow $$($$^{45}C_{8}$$+$$^{45}C_{7}$$)+$$^{46}C_{7}$$+......+$$^{51}C_{7}$$+$$\sum _{ i=1 }^{ 5 }{^{ 57-i} C_{ 7) }} $$
$$\Rightarrow $$($$^{46}C_{8}$$+$$^{46}C_{7}$$)+......+$$^{51}C_{7}$$+$$^{52}C_{7}$$+$$^{53}C_{7}$$+........+$$^{56}C_{7}$$
:
:
$$\Rightarrow$$ $$^{56}C_{8}$$+$$^{56}C_{7}$$
$$\Rightarrow$$ $$^{57}C_{8}$$

Coefficient of $${ x }^{ 15 }$$ in given expression is given by

There are total of $$n$$ brackets. The term $$x^{n-2}$$ will be formed when integers are chosen from any two brackets and $$x$$ is chosen from all the other brackets and multiplied.

Thus, the Coefficient of $$x^{n-2}$$ is 

$$ C= (1\times2+1\times3+...+1\times n)+(2\times3+2\times4+..+2\times n)+...+((n-1)\times n)$$

$$ = \left(\dfrac {(n)(n+1)}{2}-1\right)+2\left(\dfrac {(n)(n+1)}{2}-1-2\right)+...+(n-1)\left(\dfrac {(n)(n+1)}{2}-(1+2+3+..+(n-1))\right) $$

$$ = \left\{(1+2+3+...+(n-1))(\dfrac {(n)(n+1)}{2})\right\} - \left\{ 1+2(1+2)+3(1+2+3)+...+(n-1)(1+...+n-1)\right\}$$

$$ = \left\{(\dfrac {(n-1)(n)}{2})(\dfrac {(n)(n+1)}{2})\right\} - \left\{\sum_{1}^{n-1} k(\dfrac {(k)(k+1)}{2})\right\}$$

$$= \left\{ \dfrac {n^{2}(n^{2}-1)}{4}\right\} - \left\{ \sum_{1}^{n-1} \dfrac {k^{3}+k^{2}}{2} \right\} $$

$$ = \left\{ \dfrac {n^{2}(n^{2}-1)}{4}\right\} - \dfrac {1}{2}\left\{ (\dfrac {(n-1)(n)}{2})^{2} + \dfrac {(n-1)n(2n-1)}{6} \right\} $$

$$ = \dfrac {n(n-1)}{4} \left\{ {n(n+1)} -  \dfrac {n(n-1)}{2} - \dfrac {2n-1}{3} \right\} $$

$$ = \dfrac {n(n-1)}{4}\left\{ \dfrac {n(n+3)}{2} -\dfrac {2n-1}{3} \right\} $$

$$ = \dfrac {n(n-1)}{4}\left\{ \dfrac {3n^{2}+9n-4n+2}{6} \right\} $$

$$ = \dfrac {n(n-1)}{4}\left\{ \dfrac {(3n+2)(n+1)}{6} \right\} $$

$$ \therefore $$ Option $$B$$ is correct.

Writing the general term, we get 
$$T_{r}=r.\dfrac{\:^{n}C_{r}}{\:^{n}C_{r-1}}$$
$$=r.\dfrac{\dfrac{n!}{(n-r)!.r!}}{\dfrac{n!}{(n-(r-1))!.(r-1)!}}$$
$$=r.\dfrac{n-(r-1))!.(r-1)!}{(n-r)!.r!}$$
$$=r.\dfrac{(n-(r-1)}{r}$$
$$=(n-(r-1))$$
$$=n+1-r$$
Applying summation
$$\sum _{r=1} r^{n} (n+1)-r$$
$$=n(n+1)-\dfrac{n(n+1)}{2}$$
$$=(n+1)(n-\dfrac{n}{2})$$
$$=\dfrac{n(n+1)}{2}$$.