In the expansion of `(3^(-x//4)+3^(5x//4))^n` the sum of binomial coefficient is 64 and term with the greatest binomial coefficient exceeds the third

Since `(n+2)` th term is the middle term in the expansion of `(1+x)^(2n+n)`, we have `alpha = .^(2n+2)C_(n+1)`. Since `(n+1)` th and `(n+)` th terms are middle term in the...

The greatest coefficient is equal to the greatest term when `x = 1`. For `x = 1, (T_(r+1))/(T_(r)) (15-r+1)/(r ) xx 2/3` Let `(T_(r+1))/(T_(r)) ge 1` or `(15-r+1)/(r ) 2/3...

Correct Answer - `.^(7)C_(3)` or `.^(7)C_(4)` Sum of the coefficient in the expansion of `(x-2y+3z)^(n)` is `(1-2+3)^(2) = 2^(n)` (putting `x = y = z = 1`) `:. 2^(n) = 128`...

Correct Answer - `x gt (n)/(n+2)` and `x lt (n+2)/(n)` If n is even, the greatest coefficient is `.^(n)c_(n//2)`. Therefore, the greatest term is `.^(n)C_(n//2)x^(n//2)`. `:. .^(n)C_(n//2)x^(n//2)gt.^(n)C_((n//2)-1)x^((n-2)//2)` and `.^(n)C_(n//2)x^(n//2) gt .^(n)C_((n//2)+1)x^((n//2)+1)`...

Correct Answer - C Middle term of `(1+alphax)^(4)` is `T_(3)`. Its coefficient is `.^(4)C_(2)(alpha)^(2)=6alpha^(2)`. Middle term of `(1-alphax)^(6)` is `T_(4)`. Its coefficient is `.^(6)C_(3) (-alpha)^(3) = -20alpha^(3)`. According to question. `6alpha^(2)...

Correct Answer - A::B::C::D Let `T_(5)` is numerically the greatest term in the expansion of`(1+x//3)^(10)` Then, `|(T_(5))/(T_(4))| ge 1` and `|(T_(6))/(T_(5))| le 1` Now, `(T_(r+1))/(T_(r)) = (10-r+1)/(r ) x/3` `rArr |(7)/(4)...

Correct Answer - 6 `T_(r+1) = .^(n)C_(r)(x^(2))^(n-r)(-1)^(r ) x^(-r)` `= .^(n)C_(r) x^(2n-3r)(-1)^(r)` Constant term `= .^(n)C_(r)(-1)^(r)` if `2n = 3r` i.e, coefficient of `x = 0` Hence, `.^(n)C_(2pi//3)(-1)^(2n//3) = 15 = .^(6)C_(4)`

Correct Answer - B `(b)` Sequence is `t_(1)+t_(2)+t_(3)+t_(4)+…` `t_(3)=t_(1)+t_(2)`, `t_(7)=1000` `t_(1)=1` but `t_(7)=t_(1)+t_(2)+t_(3)+t_(4)+t_(5)+t_(6)` `1000=2(t_(1)+t_(2)+t_(3)+t_(4)+t_(5))` `=4(t_(1)+t_(2)+t_(3)+t_(4))` `=8(t_(1)+t_(2)+t_(3))` `=16(t_(1)+t_(2))` `:.t_(1)+t_(2)=125//2` `:. t_(2)=125//2-1=123//2`

Correct Answer - B `(b)` Sum of coefficient of `(q+r)^(20)(1+(p-2)x)^(20)` `=(q+r)^(20)(p-1)^(20)` [put `x=1`] Square of the sum of coefficient of `(2rpx-(r+q)*y^(10)` `=(2rq-(r+q))^(20)` [put `x=y=1`] So `(q+r)^(20)(p-1)^(20)=(2rq-(q+r))^(20)` `impliesp-1=(2rq)/(r+q)-1` `impliesp=(2rq)/(r+q)` `impliesp=H.M.` of `r`...

Correct Answer - 19710