Let `P = sum_(r=1)^(50)(""^(50+r)C_(r)(2r-1))/(""^(50)C_(r)(50+r)), Q = sum_(r=0)^()(""^(50)C_(r))^(2), R = sum_(0)^(100)(-1)^(r)(""^(100)C_(r))^(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 - C The given sigma is the expansion of `[(x-3)+2]^(100)=(x-1)^(100)= (1-x)^(100)` Therefore, `x^(53)`will occur in `T_(54)`. Therefore, the coefficient is `-.^(100)C_(53)`.

Correct Answer - D `(1-x)^(n)(1+x)^(n)=underset(r=0)overset(n)suma_(r)x^(r)(1-x)^(n)(1-x)^(n-r)` or `(1-x+2x)^(n) = underset(r=0)overset(n)suma_(r)x^(r)(1-x)^(n-r)` or `underset(r=0)overset(n)sum.^(n)C_(r)(1-x)^(n-r)(2x)^(r)=underset(r=0)overset(n)suma_(r)x^(r)(1-x)^(n-r)` Comparing general term, we get `a_(r) = .^(n)C_(r)2^(r)`.

Correct Answer - B In P, general term of the series is `T_(r)=(.^(50-r)C_(r)(2r-1))/(.^(50)C_(r)(50+r))` `=(.^(50+r)C_(r))/(.^(50)C_(r))(1-(50-r+1)/(50+r))` `= (.^(50+r)C_(r))/(.^(50)C_(r))-(.^(50+r)C_(r))/(.^(50)C_(r))((50-r+1)/(50+r))` Now, `(.^(50+r)C_(r))/(.^(50)C_(r))((50-r+1)/(50+r))` `= ((50-r+1)(50+r)!r!(50-r)!)/(r!50!(50+r)50!)` `=((50-r+1)!(50+r-1)!)/(50!50!)` `= (.^(50+r-1)C_(r-1))/(.^(50)C_(r-1))` `rArr T(r) = (.^(50+r)C_(r))/(.^(50)C_(r)) - (.^(50+r-1)C_(r-1))/(.^(50)C_(r-1))= V(r) -...

Correct Answer - C In P, general term of the series is `T_(r)=(.^(50-r)C_(r)(2r-1))/(.^(50)C_(r)(50+r))` `=(.^(50+r)C_(r))/(.^(50)C_(r))(1-(50-r+1)/(50+r))` `= (.^(50+r)C_(r))/(.^(50)C_(r))-(.^(50+r)C_(r))/(.^(50)C_(r))((50-r+1)/(50+r))` Now, `(.^(50+r)C_(r))/(.^(50)C_(r))((50-r+1)/(50+r))` `= ((50-r+1)(50+r)!r!(50-r)!)/(r!50!(50+r)50!)` `=((50-r+1)!(50+r-1)!)/(50!50!)` `= (.^(50+r-1)C_(r-1))/(.^(50)C_(r-1))` `rArr T(r) = (.^(50+r)C_(r))/(.^(50)C_(r)) - (.^(50+r-1)C_(r-1))/(.^(50)C_(r-1))= V(r) -...

Correct Answer - B `S_(1)=underset(j=1)overset(10)sumj(j-1)(10!)/(j(j-1)(j-2)!(10-j)!)` `= 90underset(j=1)overset(10)sum(8!)/((j-2)!(8-(j-2)!)` `=90underset(j=2)overset(10)sum.^(8)C_(j-2)=90xx2^(8)` `S_(1) = underset(j=1)overset(10)sum(10!)/(j(j-1)!(9-(j-1))!)` `= 10underset(j=1)overset(10)sum(9!)/((j-1)!(9-(j-1))!)` `10underset( j=1)overset(10)sum.^(9)C_(j-1)= 10 xx 2^(9)` `S_(3) = underset(j=1)overset(10)sum[j(j-1)+j] .^(10)C_(j)` `= underset(j=1)overset(10)sumj(j-1).^(10)C_(j)+underset(j=1)overset(10)sum..^(10)C_(j)` `= 90xx2^(8)+10xx2^(9) = 55 xx 2^(9)`

Correct Answer - B::C All are infinte geometric progression with common ratio lt 1 `x=1/(1-cos^2phi)=1/sin^2phi,y=1/(1-sin^2phi)=1/cos^2phi`, `z=1/(1-cos^2phisin^2phi)` Now, `xy+z=1/(sin^2phicos^2phi)+1/(1-sin^2phicos^2phi)` `=1/(sin^2phicos^2phi(1-sin^2phicos^2phi))` `or xy+z=xyz ...(i)` Clearly, `x+y=(sin^2phi+cos^2phi)/(sin^2phicos^2phi)=xy` `:. x+y+z=xyz` [using Eq. (i)]

Correct Answer - A `(a)` `S=sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i+j+k))`, `|a| gt 1` or `0 lt (1)/(|a|) lt 1` `=sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i)a^(j)a^(k))` `=(sum_(i=1)^(oo)(1)/(a^(i)))(sum_(j=1)^(oo)(1)/(a^(j)))(sum_(k=1)^(oo)(1)/(a^(k)))` `=((1)/(a))/(1-(1)/(a))*((1)/(a))/(1-(1)/(a))*((1)/(a))/(1-(1)/(a))=(1)/((a-1)^(3))`

Correct Answer - C `(c )` `sum_(i=1)^(n)sum_(j=1)^(i)sum_(k=1)^(j)1` `=sum_(i=1)^(n)sum_(j=1)^(i)j` `=sum_(i=1)^(n)(i(i+1))/(2)` `=(1)/(2)sum_(i=1)^(n)(i^(2)+i)` `=(n(n+1)(n+2))/(6)=220` `:.n=10`

Correct Answer - `334`