1
AIEEE 2009
+4
-1
The remainder left out when $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ is divided by 9 is :
A
2
B
7
C
8
D
0
2
AIEEE 2008
+4
-1
Out of Syllabus
Statement - 1 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r} = \left( {n + 2} \right){2^{n - 1}}.}$$
Statement - 2 : $$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r} = {{\left( {1 + x} \right)}^n} + nx{{\left( {1 + x} \right)}^{n - 1}}.}$$
A
Statement - 1 is false, Statement - 2 is true
B
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
C
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
D
Statement - 1 is true, Statement - 2 is false
3
AIEEE 2007
+4
-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
A
$$0$$
B
$${}^{20}{C_{10}}$$
C
$$- {}^{20}{C_{10}}$$
D
$${1 \over 2}{}^{20}{C_{10}}$$
4
AIEEE 2007
+4
-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
A
$${{n - 5} \over 6}$$
B
$${{n - 4} \over 5}$$
C
$${5 \over {n - 4}}$$
D
$${6 \over {n - 5}}$$
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