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JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Sum of the series ∑k=0n(2n)!(k!)2(n−k!)2\sum\limits_{{k=0}}^n \frac{(2 n) !}{(k !)^2(n-k !)^2} equals
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Correct option: B
(2nCn)2\left({ }^{2 n} C_n\right)^2
Option B ((2nCn)2\left({ }^{2 n} C_n\right)^2) is correct because the given sum can be simplified to (2nCn)∑k=0n(nCk)2\left({ }^{2 n} C_n\right) \sum_{k=0}^n\left({ }^n C_k\right)^2, which equals (2nCn)2\left({ }^{2 n} C_n\right)^2.

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(10C0)2−(10C1)2+…−(10C9)2+(10C10)2\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2 equals If aa and dd are two complex numbers, then the sum to (n+1)\left( {n + 1} \right) terms of the following series aC0−(a+d)C1+(a+2d)C2−...+...a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ... is The number of different terms in the expansion of (1−x)201(1+x+x2)200(1-x)^{201}\left(1+x+x^2\right)^{200} is The sum of the coefficient of the binomial expansion of (1x+2x)n\left(\frac{1}{x}+2 x\right)^n is equal to 6561 . The constant term in the expansion is The coefficient of x5{x^5} in the expansion of 1+x21+x,∣x∣<1,\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1, is If (2n+10)+(2n+13)+(2n+16)+…=170\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170, then nn equals
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