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1

WB JEE 2010

MCQ (Single Correct Answer)

Sum of the last 30 coefficients in the expansion of (1 + x)59, when expanded in ascending powers of x is

A
259
B
258
C
230
D
229

Explanation

Sum of last 30 coefficients in the expansion of (1 + x)59

$$ = {}^{59}{C_{30}} + {}^{59}{C_{31}} + \,\,....\,\, + {}^{59}{C_{59}}$$

$$ = {1 \over 2}[({}^{59}{C_{30}} + {}^{59}{C_{31}} + {}^{59}{C_{32}} + \,\,....\,\, + {}^{59}{C_{59}}) + ({}^{59}{C_{30}} + {}^{59}{C_{31}} + \,\,...\,\, + {}^{59}{C_{59}})]$$

$$ = {1 \over 2}[({}^{59}{C_{29}} + {}^{59}{C_{28}} + \,\,....\,\,{}^{59}{C_0}) + ({}^{59}{C_{30}} + {}^{59}{C_{31}} + \,\,....\,\, + {}^{59}{C_{59}})]$$ [$$\because$$ $${}^n{C_r} = {}^n{C_{n - r}}$$]

$$ = {1 \over 2}[{}^{59}{C_0} + {}^{59}{C_1} + \,\,....\,\, + {}^{59}{C_{29}} + {}^{59}{C_{30}} + \,\,....\,\,{}^{59}{C_{59}}]$$

$$ = {1 \over 2}{(1 + 1)^{59}} = {1 \over 2}\,.\,{2^{59}} = {2^{58}}$$

2

WB JEE 2010

MCQ (Single Correct Answer)

If in the expansion (a $$-$$ 2b)n, the sum of the 5th and 6th term is zero, then the value of $${a \over b}$$ is

A
$${{n - 4} \over 5}$$
B
$${{2(n - 4)} \over 5}$$
C
$${5 \over {n - 4}}$$
D
$${5 \over {2(n - 4)}}$$

Explanation

$$\because$$ $${t_5} + {t_6} = 0$$

$$\therefore$$ $${}^n{C_4}{a^{n - 4}}{( - 2b)^4} + {}^n{C_5}{a^{n - 5}}{( - 2b)^5} = 0$$

$$ \Rightarrow {{\left| \!{\underline {\, n \,}} \right. } \over {\left| \!{\underline {\, 4 \,}} \right. \left| \!{\underline {\, {n - 4} \,}} \right. }}a = {{\left| \!{\underline {\, n \,}} \right. } \over {\left| \!{\underline {\, 5 \,}} \right. \left| \!{\underline {\, {n - 5} \,}} \right. }}\,.\,(2b) \Rightarrow {a \over {n - 4}} = {{2b} \over 5}$$

$$ \Rightarrow {a \over b} = {{2(n - 4)} \over 5}$$

3

WB JEE 2010

MCQ (Single Correct Answer)

$$({2^{3n}} - 1)$$ will be divisible by $$(\forall n \in N)$$

A
25
B
8
C
7
D
3

Explanation

$$n \in N$$ $$\therefore$$ n = 1, 2, 3, ......

$$\therefore$$ Putting n = 1, $${2^{3n}} - 1 = {2^3} - 1 = 7$$

$$\therefore$$ $${2^{3n}} - 1$$ will be divisible by $$7\forall n \in N$$

4

WB JEE 2009

MCQ (Single Correct Answer)

using binomial theorem, the value of (0.999)3 correct to 3 decimal places is

A
0.999
B
0.998
C
0.997
D
0.995

Explanation

(0.999)3 = (1 $$-$$ 0.001)3

= 1 $$-$$ 3 $$\times$$ (0.001) = 0.997.

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