The value of `sum_(i=1)^(n)sum_(j=1)^(i)sum_(k=1)^(j)=220` , then the value of `n` equals

Correct Answer - `8/3` `lim_(n to oo) (sum_(r=1)^(n)sqrt(r) sum_(r=1)^(n)1/(sqrt(4)))/(sum_(r=1)^(n)r)` `:.` Limit `=lim_(nto oo) (1/n sum_(r=1)^(n)sqrt(4/n)(1/nsum_(r=1)^(n)sqrt(n/r)))/(1/n sum_(r=1)^(n)r/n)` `=(int_(0)^(1)sqrt(x)dxint_(0)^(1)(dx)/(sqrt(x)))/(int_(0)^(1)x dx) =8/3`

`underset(r=0)overset(n)sum(-1)^(r).a_(r )..^(n)C_(r )` `= a_(0) xx .^(n)C_(0) - a_(1) xx .^(n)C_(1) + a_(2) xx .^(n)C_(2)-"….."+(-1)^(n)a_(n) xx .^(n)C_(0)` `=` Coefficient of `x^(n)` in `(1-x+x^(2))^(n)(x-1)^(n)` `=` Coefficient of `x^(n)` in `(x^(3)-1)^(n)` `=...

Correct Answer - `(3^(n) - 2^(n))` and `1` `S = underset(p=1)overset(n)sum(underset(m=p)overset(n)sum.^(n)C_(m).^(m)C_(p))` `= underset(p=1)overset(n)sum(.^(n)C_(p).^(p)C_(p)+.^(n)C_(p+1).^(p)+1)C_(p) + "....."+.^(n)C_(n).^(n)C_(p))` `= underset(p=1)overset(n)sum` [coefficient of `x^(p)` in `{.^(n)C_(p)(1+x)^(p)+.^(n)C_(p+1)(1+x)^(p+1)+"....."+.^(n)C_(n)(1+x)^(n)}]` `= underset(p=1)overset(n)sum` [Coefficeint of `x^(p)` in `{.^(n)C_(0)+.^(n)C_(1)(1+x)+.^(n)C_(2)(1+x)^(2)+"....."+.^(n)C_(p-1)(1+x)^(p-1)+.^(n)C_(p)(1+x)^(p)+"....."+.^(n)C_(n)(1+x)^(n)}]` `= underset(p=1)oversetg(n)sum`...

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 - 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::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 - C `(c )` `S=sum_(n=0)^(100)(i)^(n!)` `=(i)^(0!)+(i)^(1!)+(i)^(2!)+…..` `=i+i-1+i^(6)+i^(24)+(i)^(5!)+(i)^(6!)+..+(i)^(100!)` `=95+2i`

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 - `334`

Correct Answer - `ubrace(333"......"3)_("n times")`