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1

### AIEEE 2008

MCQ (Single Correct Answer)
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

## Explanation

Check Statement - 1

$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}}$$

= $$\sum\limits_{r = 0}^n {r.{}^n{C_r}}$$ + $$\sum\limits_{r = 0}^n {{}^n{C_r}}$$

= $$\sum\limits_{r = 1}^n {r.{n \over r}{}^{n - 1}{C_{r - 1}}}$$ $$+ {2^n}$$

= $$n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}}$$ $$+ {2^n}$$

= $$n \times {2^{n - 1}}$$$$+ {2^n}$$

= $${2^{n - 1}}\left[ {n + 2} \right]$$

Check Statement 2 :

$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{}^n{C_r}{x^r}}$$

= $$\sum\limits_{r = 0}^n r .{}^n{C_r}.{x^r}$$ + $$\sum\limits_{r = 0}^n {{}^n{C_r}.{x^r}}$$

= $$n\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^r}}$$ $$+ {\left( {1 + x} \right)^n}$$

= $$nx\sum\limits_{r = 1}^n {{}^{n - 1}{C_{r - 1}}.{x^{r - 1}}} + {\left( {1 + x} \right)^n}$$

= $$nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n}$$

Substitude x = 1 in the statement 2 and we get,

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

So Option B is correct.
2

### AIEEE 2007

MCQ (Single Correct Answer)
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
$${6 \over {n - 5}}$$

## Explanation

According to the question,

t5 + t6 = 0

$$\therefore$$ $${}^n{C_4}.{a^{n - 4}}.{b^4}$$ + $$\left( { - {}^n{C_5}.{a^{n - 5}}.{b^5}} \right)$$ = 0

By solving we get,

$${a \over b} = {{n - 4} \over 5}$$
3

### AIEEE 2007

MCQ (Single Correct Answer)
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
$${1 \over 2}{}^{20}{C_{10}}$$

## Explanation

We know

$${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... + {}^{20}{C_{10}} - {}^{20}{C_{11}}+ ...... + {}^{20}{C_{20}} = 0$$

$$\Rightarrow$$ $$({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$$ $$+{}^{20}{C_{10}}$$ $$(-{}^{20}{C_{9}}+ {}^{20}{C_{8}}+...... + {}^{20}{C_{0}})$$

(As $${}^{20}{C_{11}} = {}^{20}{C_9}$$)

$$\Rightarrow$$ $$2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}})$$ $$+{}^{20}{C_{10}}$$ = 0

$$\Rightarrow$$ $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$$ = $$- {1 \over 2}{}^{20}{C_{10}}$$

Adding $${}^{20}{C_{10}}$$ both sides,

$$\Rightarrow$$ $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$$ $$+ {}^{20}{C_{10}}$$ = $$- {1 \over 2}{}^{20}{C_{10}}$$ $$+ {}^{20}{C_{10}}$$

$$\Rightarrow$$ $${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + .... - {}^{20} {C_{9}}$$ $$+ {}^{20}{C_{10}}$$ = $${1 \over 2}{}^{20}{C_{10}}$$
4

### AIEEE 2006

MCQ (Single Correct Answer)
For natural numbers $$m$$ , $$n$$, if $${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + ..........$$ and $${a_1} = {a_2} = 10,$$ then $$\left( {m,\,n} \right)$$ is
A
$$\left( {20,\,45} \right)$$
B
$$\left( {35,\,20} \right)$$
C
$$\left( {45,\,35} \right)$$
D
$$\left( {35,\,45} \right)$$

## Explanation

$${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\,$$

= $$\left( {{}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + ....} \right)$$ -
$$\left( {{}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + ....} \right)$$

$${a_1}$$ = Coefficient of y = $${{}^n{C_1}}$$ - $${{}^m{C_1}}$$ = 10

$$\Rightarrow$$ n - m = 10

$${a_2}$$ = Coefficient of y2

= $${}^n{C_2} + {}^n{C_1} \times {}^m{C_1} + {}^m{C_2} = 10$$

$$\Rightarrow {{n\left( {n - 1} \right)} \over 2} - nm + {{m\left( {m - 1} \right)} \over 2} = 10$$

$$\Rightarrow n\left( {n - 1} \right) - 2nm + m\left( {m - 1} \right) = 20$$

$$\Rightarrow$$ (m + 10)(m + 9) - 2(m + 10)m + m(m - 1) = 20

$$\Rightarrow$$ 90 + 19m + m2 - 2m2 - 20m + m2 - m - 20 = 0

$$\Rightarrow$$ 70 - 2m = 0

$$\Rightarrow$$ m = 35

$$\therefore$$ n = 10 + 35 = 45

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