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1

AIEEE 2003

MCQ (Single Correct Answer)
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
2

AIEEE 2003

MCQ (Single Correct Answer)
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.
3

AIEEE 2002

MCQ (Single Correct Answer)
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 $${a_n} > 3\,\,\forall \,\,n \ge 1$$
4

AIEEE 2002

MCQ (Single Correct Answer)
If the sum of the coefficients in the expansion of $$\,{\left( {a + b} \right)^n}$$ is 4096, then the greatest coefficient in the expansion is
A
1594
B
792
C
924
D
2924

Explanation

We know, $$\,{\left( {a + b} \right)^n}$$ = $${}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}$$

Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable.

So put $$a$$ = b = 1

$$\therefore$$ 2n = $${}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}$$

According to question, 2n = 4096 = 212

$$ \Rightarrow n = 12$$

So $$\,{\left( {a + b} \right)^n}$$ = $$\,{\left( {a + b} \right)^{12}}$$

Here n = 12 is even so formula for greatest term is
$${T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}$$

For n = 12, greatest term $${T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}$$

$$\therefore$$ Coefficient of the greatest term = $${}^{12}{C_6}$$ = $${{12!} \over {6!6!}}$$ = 924

Questions Asked from Mathematical Induction and Binomial Theorem

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