Prove that `1/(n+1)=(^n C_1)/2-(2(^n C_2))/3+(3(^n C_3))/4-+(-1)^(n+1)(n(^n C_n))/(n+1)` .

Correct Answer - `f(x) = x^(3) - x^(2)`

`S = C_(1)(a-1)- C_(2)(a-2) + "…." + (-1)^(n-1)C_(n)(a-n)` `:. T_(r) = (-1)^(r-1)(a-r).^(n)C_(r)` `= (-1)^(r-1)(a.^(n)C_(r) - r.^(n)C_(r))` `= (-1)^(r-1)(a.^(n)C_(r)-n.^(n-1)C_(r-1))` ` = - a (-1)^(r ). .^(n)C_(r) - n (-1)^(r=1 xx n...

`.^(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)`

To find `S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)` Consider, `.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)` `= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)` = Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)` `= .^(200)C_(98) " "(1)` Also,...

Correct Answer - `1` We have `(1)/(81^(n)) - (10)/(81^(n)).^(2n)C_(1)+(10^(2))/(81^(n)).^(2n)C_(2)- (10^(3))/(81^(n)) .^(2n)C_(3)+"……."+(10^(2n))/(81^(n))` `= 1/(81^(n))[.^(2n)C_(0) - .^(2n)C_(1)10^(1) + .^(2n)C_(2)10^(2)-.^(2n)C_(3)10^(3)+"……"+.^(2n)C_(2n)10^(2n)]` `= (1)/(81^(n))[1-10]^(2n)` `= ((-9)^(2n))/(81^(n))= (81^(n))/(81^(n)) = 1`

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) =...

Correct Answer - C `(1+x)^(n)=C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+.^(n)C_(3)x^(3)+"....."+C_(n-1)x^(n-1)+C_(n)x^(n) " "(1)` `(x+1)^(n) = C_(0)x^(n)+C_(1)x^(n-1)+C_(2)x^(n-2)+"...."+C_(n-1)x+C_(n)" "(2)` Multiplying Eqs. (1) and (2) and equating the coefficient of `x^(n-2)`. we get `C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+"....."C_(n-2)C_(n)` `=` coefficient of `x^(n-2)` in `(1+x)^(2n)`...

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) +...

Correct Answer - A::B::D `((n-1)(n-2)"……"(n-m+1))/((m-1)!)` = `=((n-1)(n-2)"...."(n-m+1)(n-m)"...."2.1)/((n-m)!(m-1)!)` `=.^(n-1)C_(m-1)` `=` Coefficient of `x^(m-1)` in `(1+x)^(n-1)` `=` Coefficient of `x^(m-1)` in `(1+x)^(n)(1+x)^(-1)` Now, `(1+x)^(n) = C_(0)+C_(1)x+C_(2)x^(2)+"...."+C_(m-1)x^(m-1)+"...."+C_(n)x^(n)" "(1)` `(1+x)^(-1) = 1-x+x^(2)-x^(3)+"...."+(-1)^(m-1)x^(m-1)+"......"" "(2)` Collecting the...

Correct Answer - Inconsistent