If the sum of the coefficients in the expansion of (1 + 2x)^n is 6561, the greatest term in the expansion for x = 1/2 is

Q: If the sum of the coefficients in the expansion of (1 + 2x)n is 6561, the greatest term in the expansion for x = 1/2 is

(A) 4th

(B) 5th

(C) 6th

(D) none of these

Click to See Answer :
Sol. sum of the coefficient in the expansion of (1 + 2x)n = 6561

⇒ (1 + 2x)n = 6561, when x = 1

⇒ 3n = 6561

⇒ 3n = 38

⇒ n = 8

$\large \frac{T_{r+1}}{T_r} = \frac{8_{C_r} (2x)^r }{8_{C_{r-1}} (2x)^{r-1}}  = \frac{9-r}{r} \times 2x $

Since x = 1/2 ;

$\large \frac{T_{r+1}}{T_r} = \frac{9-r}{r}  $

$\large \frac{T_{r+1}}{T_r} > 1 \Rightarrow  \frac{9-r}{r} >1  $

9-r > r  ⇒ 2r < 9 ⇒ r < 4.5

Hence, 5th term is the greatest term.

Hence (B) is the correct answer.

 

The digits at unit’s place in the number $ 17^{1995} + 11^{1995} – 7^{1995}$ is

Q: The digits at unit’s place in the number $\large 17^{1995} + 11^{1995} – 7^{1995}$ is

(A) 0

(B) 1

(C) 2

(D) 3

Sol: $\large 17^{1995} + 11^{1995} – 7^{1995}$

$\large = (7+10)^{1995} + (1+10)^{1995} – 7^{1995}$

$\large = [ 7^{1995} + 1995_{C_1} 7^{1994} 10^1 + 1995_{C_2} 7^{1993} 10^2 + ……+ 1995_{C_{1995}} 10^{1995}] $

$\large + [ 1995_{C_0} + 1995_{C_1} 10^1 + 1995_{C_2} 10^2 + ……+ 1995_{C_{1995}} 10^{1995} ] -7^{1995}$

$\large [ 1995_{C_1} 7^{1994} 10^1 + ….+ 10^{1995}] $ $\large + [1995_{C_1} 10^1 + ….+ 1995_{C_{1995}} 10^{1995} ] + 1 $

= a multiple of 10 + 1

Thus, the unit’s place digits is 1

Hence (B) is the correct answer.

The coefficient of x^5 in the expansion of $ (1 + x)^{21} +(1+x)^{22} +…+ (1 + x)^{30} $ is

Q: The coefficient of x5 in the expansion of $\large (1 + x)^{21} +(1+x)^{22} +…+ (1 + x)^{30} $ is

(A) 51C5

(B) 9C5

(C) 31C621C6

(D) 30C5 + 20C5

Sol: $\large (1 + x)^{21} +(1+x)^{22} +…+ (1 + x)^{30} $

$\large = (1 + x)^{21} (\frac{(1+x)^{10}-1}{(1+x)-1})$

$\large = \frac{1}{x} [(1+x)^{31} – (1+x)^{21} ] $

coefficient of x5 in the given expression

$\large = coefficient \; of \; x^5 in \; \frac{1}{x} [(1+x)^{31} – (1+x)^{21} ] $

$\large = coefficient \; of \; x^6 in \; [(1+x)^{31} – (1+x)^{21} ] $

$\large = 31_{C_6} – 21_{C_6} $

Hence (C) is the correct answer.

The value of $ 99^{50} – 99.98^{50} + \frac{99 . 98}{1.2} (97)^{50} – ……+ 99 $ is

Q: The value of $\displaystyle 99^{50} – 99.98^{50} + \frac{99 . 98}{1.2} (97)^{50} – ……+ 99 $ is

(A) 0

(B) – 1

(C) – 2

(D) – 3

Sol: $\displaystyle 99^{50} – 99.98^{50} + \frac{99 . 98}{1.2} (97)^{50} – ……+ 99 $

$\displaystyle = 99^{50} – 99_{C_1}(98)^{50} + 99_{C_2} (97)^{50} – ……+ 99_{C_{98}}.1 $

$\displaystyle = 99_{C_0} 99^{50} – 99_{C_1}(99-1)^{50} + 99_{C_2} (99-2)^{50} – ……$

$\displaystyle + 99_{C_{98}} (99-98)^{50} -99_{C_{99}} (99-99)^{50}$

$\displaystyle = 99^{50}( 99_{C_0} – 99_{C_1} + 99_{C_2} – …..+ 99_{C_{98}} -99_{C_{99}} ) $

$\displaystyle + 50_{C_1} 99^{49}( 99_{C_1} – 2 . 99_{C_2} + 3 . 99_{C_3} – …… )+ … $

= 0 + 0 +… = 0

Hence (A) is the correct answer.