Prove that `^100 C_2^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[^(200)C_(98)-^(100)C_(49)]dot`

Correct Answer - `(3-sqrt(5))/2` `int_(-1)^(1)[x^(2)+{x}]dx=int_(-1)^(0)[x^(2)+x+1]dx+int_(0)^(1)[x^(2)+x]dx` `=0+int_(0)^((sqrt(5)-1)/2)0dx+int_((sqrt(5)-1)/2)^(1)1dx` `=(3-sqrt(5))/2`

`.^(m)C_(4)-.^(m+n)C_(1).^(m)C_(3)+.^(m+n)C_(2).^(m)C_(2)-.^(m+n)C_(3).^(m)C_(1)+.^(m+n)C_(4)` `= .^(m+n)C_(0).^(m)C_(4)-.^(m+n)C_(1).^(m)C_(3)+.^(m+n)C_(2).^(m)C_(2)-.^(m+n)C_(3).^(m)C_(1)+.^(m+n)C_(4).^(m)C_(0)` = Coefficient of `x^(4)` in `(1+x)^(m+n)(1-x)^(m)` = Coefficient of `x^(4)` in `(1-x)^(m)(1+x)^(n)` = Coefficient of `x^(4)` in `[1-.^(m)C_(1)x^(2)+.^(m)C_(2)x^(4)-"....."][1+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)]` `= .^(n)C_(4)-.^(m)C_(1) xx .^(n)C_(2)+.^(m)C_(2)`

`(18x^(2)+12x+4)^(n) = 2^(n)[2+9x^(2)+6x]^(n)` Now, `a_(r )` is coefficient of `x^(r )` in `2^(n) [(3x+1)^(2)+1]^(n)`. Hence `a_(r) =` Coefficient of `x^(r )2^(n)[.^(n)C_(0)(3x+1)^(2n)+.^(n)C_(1)(3x+1)^(2n-2) + .^(n)C_(2)(3x+1)^(2n-4)+"…."+.^(n)C_(r )(3x+1)^(2n-2r)+"....."]` or `a_(r)=2^(n)[.^(n)C_(0)3^(r).^(2n)C_(r)+.^(n)C_(1)3^(r).^(2n-2)C_(r)+.^(n)C_(2)3^(r).^(2n-4)C_(r)+"...."]` `= 2^(n)3^(r)[.^(n)C_(0).^(2n)C_(r)+.^(n)C_(1).^(2n-2)C_(r)+.^(n)C_(2).^(2n-4)C_(r)+"...."]`

Correct Answer - `(n+1)2^(n)` Given polynomial is `n +1` degree polynomial. `(x+.^(n)C_(0))(x+3.^(n)C_(2))(x+5.^(n)C_(2))"....."[x+(2n+1).^(n)C_(n)]` `= x^(n+1)+a_(1)x^(n)+"...."+a_(n+1)` Then, `- (a_(1))/(a_(0))=-.^(n)C_(0)-3.^(n)C_(1)-5.^(n)C_(2)-"....."-(2n+1).^(n)C_(n)` `rArr a_(1)=.^(n)C_(0)+3.^(n)C_(1)+5.^(n)C_(2)-"......"+(2n+1).^(n)C_(n)` `= underset(r=0)overset(n)sum.^(n)C_(r)(2r+1)=2underset(r=0)overset(n)sumr^(n)C_(r)+underset(r=0)overset(n)sum.^(n)C_(r)` `= 2underset(r=0)overset(n)sumn..^(n-1)C_(r-1) + underset(r=0)overset(n)sum.^(n)C_(r)` `= 2n 2^(n+1) + 2^(n) =...

`S = (.^(n)C_(1))/(2)-(2(.^(n)C_(3)))/(3)+(3(.^(n)C_(3)))/(4)"...."+(-1)^(n+1)(n(.^(n)C_(n)))/(n+1)` ` = underset(r=1)overset(n)sum (r.^(n)C_(r))/((r+1))(-1)^(r+1)` `= underset(r=1)overset(n)sum r(.^(n+1)C(r+1))/(n+1)(-1)^(r+1)` `= 1/(n+1) underset(r=1)overset(n)sum [(r+1).^(n+1)C_(r+1)(-1)^(r+1)-.^(n+1)C_(r+1)(-1)^(r+1)]` `= (1)/(n+1)underset(r=1)overset(n)sum[-(n+1).^(n)C_(r)(-1)^(r)-.^(n+1)C_(r+1)(-1)^(r+1)]` `= -[-.^(n)C_(2)-.^(n)C_(3)+"...."+(-1)^(n).^(n)C_(n)]` ` - 1/(n+1)[.^(n+1)C_(2) - .^(n+1)C_(3)+".....(-1)^(n+1).^(n+1)C_(n+1)]` `= - [(1-1)^(n)-1]-(1)/(n+1)[(1-1)^(n+1) - .^(n+1)C_(0)+.^(n+1)C_(1)]` ` =...

Correct Answer - B `(1+x^(3)-x^(6))^(30)` `= {1+x^(3)(1-x^(3))}^(30)` `=.^(30)C_(0) + .^(30)C_(1)x^(3)(1-x^(3))+.^(30)C_(2)x^(6)(1-x^(3))^(2) + "…."` Obviously, each term will contain `x^(3m), m in N`. But 28 is not divisible by 3. Therefore ther will...

Correct Answer - A::B::D `(1+z^(2)+z^(4))^(8) = C_(0) + C_(1)z^(2) + X_(2)z^(4) + "….." + C_(16)z^(32) " "(1)` Putting `x = i`, where `i = sqrt(-1)`. `(1-1+1)^(8) = C_(0) - C_(1) +...

`(d^(n))/(dx^(n))[f(x)]=|{:((d^(n))/(dx^(n))(x^(n)),n!,2),((d^(n))/(dx^(n))(cos x),cos (npi)/(2),4),((d^(n))/(dx^(n))(sin x), sin (npi)/(2),8):}|` `=|{:(n!,n!,2),(cos(x+(npi)/(2)),cos""(npi)/(2),4),(sin(x+(npi)/(2)),sin""(npi)/(2),8):}|` `"or "(d^(n))/(dx^(n))[f(x)]_(x=0)=0`

Correct Answer - A `(a)` Writing `((200),(r ))=((200),(200-r))`, we have `((100),(0))((200),(50))+((100),(1))((200),(49))+((100),(2))((200),(48))+......+((100),(50))((200),(0))` `=` Coefficient of `x^(50)` in the expansion of `(1+x)^(100)(x+1)^(200)` `=` Coefficient of `x^(50)` in the binomial expansion of `(1+x)^(300)` `=((300),(50))`

Correct Answer - Inconsistent