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1

### AIEEE 2010

Let $${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$,

$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$$ and

$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$$

Statement-1 : $${{S_3} = 55 \times {2^9}}$$.
Statement-2 : $${{S_1} = 90 \times {2^8}}$$ and $${{S_2} = 10 \times {2^8}}$$.

A
Statement - 1 is true, Statement- 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B
Statement - 1 is true, Statement-2 is false.
C
Statement - 1 is false, Statement-2 is true.
D
Statement - 1 is true, Statement-2 is true: -Statement - 2 is a correct explanation for Statement - 1.

## Explanation

Note :

$$\sum\limits_{r = 0}^n {r.{}^n{C_r}}$$ = $$= n{.2^{n - 1}}$$

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

Given that,

$${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$

=$$\sum\limits_{j = 1}^{10} {{j^2}.{}^{10}} {C_j} - \sum\limits_{j = 1}^{10} {j.{}^{10}} {C_j}$$

= 10$$\times$$11$$\times$$$${2^{10 - 2}}$$ - 10$$\times$$$${2^{10 - 1}}$$

= 10$$\times$$$${2^{8}}$$(11 - 2)

= 10$$\times$$9$$\times$$$${2^{8}}$$

= 90$$\times$$$${2^{8}}$$

$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$$

= 10$$\times$$$${2^{10-1}}$$

= 10$$\times$$$${2^{9}}$$

$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$$

= 10$$\times$$11$$\times$$$${2^{10-2}}$$

= $${{110} \over 2} \times {2^9}$$

= 55 $$\times$$ $${2^9}$$
2

### AIEEE 2009

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

## Explanation

$${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$

= $${\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$$

= $${\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}$$

= $$\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)$$
+ $$\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)$$

= 2 + 63$$\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]$$

= 63$$\times$$[Some integral value] + 2

63$$\times$$[Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.
3

### AIEEE 2008

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.
4

### AIEEE 2007

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,

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

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