`sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i+j+k))` is equal to (where `|a| gt 1`)

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`

Correct Answer - A `underset(r=1)overset(n+1)sum(underset(k=1)overset(n)sum.^(k)C_(r-1))= underset(r=1)overset(n+1)sum(underset(k=1)overset(n)sum(.^(k+1)C_(r)-.^(k)C_(r)))` `= underset(r=1)overset(n+1)sum(.^(n+1)C_(r)+.^(1)C_(r))` `= 2^(n+1) - 2`

Correct Answer - D In the given equation, put `x - 3 = y`. `:. underset(r=0)overset(2n)suma_(r)(1+y)^(r) = underset(r=0)overset(2n)sumb_(r)(y)^(r)` `rArr a_(0) + a_(1)(1+y) + "….."+ a_(n+1)(1+y)^(n-1)+(1+y)^(n) + (1+y)^(n+1)+"…."+(1+y)^(2n)` `= underset(r=0)overset(2n)sumb_(r)y^(r)` [Using `a_(k) =...

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 )` `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 - B `(b)` `sum_(k=1)^(10)underset(i nej ne k)(sum_(j=1)^(10))sum_(i=1)^(10)1` `=`(sum when `i`, `j`, `k` are independent) `-`(sum whenany two of `i`,`j`,`k` are equal) `-`(sum when all `i`, `j`, `k` are equal)...

Correct Answer - `334`

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