AIEEE 2009 | Binomial Theorem Question 193 | Mathematics

Statement - 1 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.} $$
Statement - 2 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.} $$

Statement - 1 is false, Statement - 2 is true

Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1

Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1

Statement - 1 is true, Statement - 2 is false

AIEEE 2007

MCQ (Single Correct Answer)

-1

Out of Syllabus

The sum of the series $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .....\, - \,.....\, + {}^{20}{C_{10}}$$ is

$$0$$

$${}^{20}{C_{10}}$$

$$ - {}^{20}{C_{10}}$$

$${1 \over 2}{}^{20}{C_{10}}$$

AIEEE 2007

MCQ (Single Correct Answer)

-1

In the binomial expansion of $${\left( {a - b} \right)^n},\,\,\,n \ge 5,$$ the sum of $${5^{th}}$$ and $${6^{th}}$$ terms is zero, then $$a/b$$ equals

$${{n - 5} \over 6}$$

$${{n - 4} \over 5}$$

$${5 \over {n - 4}}$$

$${6 \over {n - 5}}$$

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