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1

### AIEEE 2004

Let $$S(K)$$ $$= 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}.$$ Then which of the following is true
A
Principle of mathematical induction can be used to prove the formula
B
$$S\left( K \right) \Rightarrow S\left( {K + 1} \right)$$
C
$$S\left( K \right) \ne S\left( {K + 1} \right)$$
D
$$S\left( 1 \right)$$ is correct

## Explanation

Given $$S(K)$$ $$= 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}$$

When k = 1, S(1): 1 = 3 + 1,

L.H.S of S(k) $$\ne$$ R.H.S of S(k)

So S(1) is not true.

As S(1) is not true so principle of mathematical induction can not be used.

S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1) = 3 + (k + 1)2

Now let S(k) is true

$$\therefore$$ 1 + 3 + 5 +........(2k - 1) = 3 + k2

$$\Rightarrow$$ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k2 + 2k +1

= 3 + (k + 1)2

$$\Rightarrow$$ S(k + 1) is true.

$$\therefore$$ S(k) $$\Rightarrow$$ S(k + 1)
2

### AIEEE 2003

The number of integral terms in the expansion of $${\left( {\sqrt 3 + \root 8 \of 5 } \right)^{256}}$$ is
A
35
B
32
C
33
D
34

## Explanation

General term = $${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$$
= $${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$$

When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.
And when $${{r \over 8}}$$ is integer then $${\left( 5 \right)^{{r \over 8}}}$$ is integer.

Entire general term will be integer when $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ both are integer.

$${{{256 - r} \over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256

$${{r \over 8}}$$ is integer when r = 0, 8, 16 ,......., 256

Now both $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, tn = a + (n - 1)d

$$\therefore$$ n = $${{256} \over 8} + 1$$ = 32 + 1 = 33
3

### AIEEE 2003

If $$x$$ is positive, the first negative term in the expansion of $${\left( {1 + x} \right)^{27/5}}$$ is
A
6th term
B
7th term
C
5th term
D
8th term.

## Explanation

General term of $${\left( {1 + x} \right)^{n}}$$ is ($${T_{r + 1}}$$) = $${{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}$$

$$\therefore$$ General term of $${\left( {1 + x} \right)^{27/5}}$$ = $${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r + 1} \right)} \over {1.2.3....r}}{x^r}$$

For first negative term, $${\left( {{{27} \over 5} - r + 1} \right)}$$ < 0

$$\Rightarrow r > {{27} \over 5} + 1$$

$$\Rightarrow r > {{32} \over 5}$$

$$\Rightarrow r > 6.4$$

$$\therefore$$ r = 7

$${T_{7 + 1}} = {T_8}$$ means 8th term is the first negative term.
4

### AIEEE 2002

If $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } }$$ having $$n$$ radical signs then by methods of mathematical induction which is true
A
$${a_n} > 7\,\,\forall \,\,n \ge 1$$
B
$${a_n} < 7\,\,\forall \,\,n \ge 1$$
C
$${a_n} < 4\,\,\forall \,\,n \ge 1$$
D
$${a_n} > 3\,\,\forall \,\,n \ge 1$$

## Explanation

Given $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } }$$

$$\therefore$$ $${a_n} = \sqrt {7 + {a_n}}$$

$$\Rightarrow$$ $$a_n^2 = 7 + {a_n}$$

$$\Rightarrow$$ $$a_n^2 - {a_n} - 7 = 0$$

$$\Rightarrow {a_n} = {{1 \pm \sqrt {1 - 4 \times 1 \times - 7} } \over 2}$$

$$\Rightarrow {a_n} = {{1 \pm \sqrt {29} } \over 2}$$

As $${a_n}$$ > 0,

$$\therefore$$ $${a_n} = {{1 + \sqrt {29} } \over 2}$$ = 3.19

So $${a_n} > 3\,\,\forall \,\,n \ge 1$$

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