If 1 + 3 x - 2 x 2 10 = a 0 + a 1 x + a 2 x 2 + … + a 20 x 20 , then prove that a 0 - a 1 + a 2 - a 3 + … + a 20 = 4 10 .

Given, 1 + 3 x - 2 x 2 10 = a 0 + a 1 x + a 2 x 2 + … + a 20 x 20 Put, x = - 1 ⇒ ( 1 - 3 - 2 ) 10 = a 0 - a 1 + a 2 - a 3 + … + a 20 - 4 10 = a 0 - a 1 + a 2 - a 3 + … + a 20 ∴ a 0 - a 1 + a 2 - a 3 + … + a 20 = 4 10 Hence, proved.

If R , n are positive integers, n is odd, 0 < F < 1 and if ( 5 5 + 11 ) n = R + F , then prove that R is an even integer.

Since R , n are positive integers, 0 < F < 1 and ( 5 5 + 11 ) n = R + F Let, ( 5 5 - 11 ) n = f Now, 11 < 5 5 < 12 ⇒ 0 < 5 5 - 11 < 1 ⇒ 0 < ( 5 5 - 11 ) n < 1 ⇒ 0 < f < 1 ⇒ 0 > - f > - 1 ∴ - 1 < - f < 0 Now, R + F - f = ( 5 5 + 11 ) n - ( 5 5 - 11 ) n = C 0 n ( 5 5 ) n + C 1 n ( 5 5 ) n - 1 ( 11 ) + C 2 n ( 5 5 ) n - 2 ( 11 ) 2 + … + C n n ( 11 ) n - C 0 n ( 5 5 ) n - C 1 n ( 5 5 ) n - 1 ( 11 ) + C 2 n ( 5 5 ) n - 2 ( 11 ) 2 + … + C n n ( - 11 ) n = 2 C 1 n ( 5 5 ) n - 1 ( 11 ) + C 3 n ( 5 5 ) n - 3 ( 11 ) 2 + … = 2 k where k is an integer ∴ R + F - f is an even integer ⇒ F - f is an integer since R is an integer. But, 0 < F < 1 and - 1 < - f < 0 ⇒ - 1 < F - f < 1 ∴ F - f = 0 ⇒ F = f ∴ R is an even integer.

If R , n are positive integers, n is odd, 0 < F < 1 and if ( 5 5 + 11 ) n = R + F , then prove that ( R + F ) · F = 4 n .

Since R , n are positive integers, 0 < F < 1 and ( 5 5 + 11 ) n = R + F Let, ( 5 5 - 11 ) n = f Now, 11 < 5 5 < 12 ⇒ 0 < 5 5 - 11 < 1 ⇒ 0 < ( 5 5 - 11 ) n < 1 ⇒ 0 < f < 1 ⇒ 0 > - f > - 1 ∴ - 1 < - f < 0 Now, R + F - f = ( 5 5 + 11 ) n - ( 5 5 - 11 ) n = C 0 n ( 5 5 ) n + C 1 n ( 5 5 ) n - 1 ( 11 ) + C 2 n ( 5 5 ) n - 2 ( 11 ) 2 + … + C n n ( 11 ) n - C 0 n ( 5 5 ) n - C 1 n ( 5 5 ) n - 1 ( 11 ) + C 2 n ( 5 5 ) n - 2 ( 11 ) 2 + … + C n n ( - 11 ) n = 2 C 1 n ( 5 5 ) n - 1 ( 11 ) + C 3 n ( 5 5 ) n - 3 ( 11 ) 3 + … = 2 k where k is an integer ∴ R + F - f is an even integer ⇒ F - f is an integer since R is an integer. But, 0 < F < 1 and - 1 < - f < 0 ⇒ - 1 < F - f < 1 ∴ F - f = 0 ⇒ F = f ∴ R is an even integer. To prove, ( R + F ) · F = 4 n LHS = ( R + F ) F = ( R + F ) f ∵ F = f = ( 5 5 + 11 ) n ( 5 5 - 11 ) n = [ ( 5 5 + 11 ) ( 5 5 - 11 ) ] n = ( 125 - 121 ) n = 4 n ∴ ( R + F ) F = 4 n Hence, proved.

If I , n are positive integers, 0 < f < 1 and if ( 7 + 4 3 ) n = I + f , then show that I is an odd integer.

Given, I , n are positive integers and ( 7 + 4 3 ) n = I + f , 0 < f < 1 Let, 7 - 4 3 n = F Now, 6 < 4 3 < 7 ⇒ - 6 > - 4 3 > - 7 ⇒ 1 > 7 - 4 3 > 0 ⇒ 0 < ( 7 - 4 3 ) n < 1 ∴ 0 < F < 1 I + f + F = ( 7 + 4 3 ) n + ( 7 - 4 3 ) n = C 0 n 7 n + C 1 n 7 n - 1 ( 4 3 ) + C 2 n 7 n - 2 ( 4 3 ) 2 + … + C n n ( 4 3 ) n + C 0 n 7 n - C 1 n 7 n - 1 ( 4 3 ) + C 2 n 7 n - 2 ( 4 3 ) 2 + … + C n n ( - 4 3 ) n = 2 C 0 n 7 n + C 2 n 7 n - 2 ( 4 3 ) 2 … … = 2 k , where k is an integer ∴ I + f + F is an even integer. ⇒ f + F is an integer since I is an integer. But 0 < f < 1 and 0 < F < 1 ⇒ f + F < 2 ∴ f + F = 1 ⇒ I + 1 is an even integer. ∴ I is an odd integer.

If I , n are positive integers, 0 < f < 1 and if ( 7 + 4 3 ) n = I + f , then show that ( I + f ) ( 1 - f ) = 1 .

Given, I , n are positive integers and ( 7 + 4 3 ) n = I + f , 0 < f < 1 Let, ( 7 - 4 3 ) n = F Now, 6 < 4 3 < 7 ⇒ - 6 > - 4 3 > - 7 ⇒ 1 > 7 - 4 3 > 0 ⇒ 0 < ( 7 - 4 3 ) n < 1 ∴ 0 < F < 1 I + f + F = ( 7 + 4 3 ) n + ( 7 - 4 3 ) n = C 0 n 7 n + C 1 n 7 n - 1 ( 4 3 ) + C 2 n 7 n - 2 ( 4 3 ) 2 + … + C n n ( 4 3 ) n + C 0 n 7 n - C 1 n 7 n - 1 ( 4 3 ) + C 2 n 7 n - 2 ( 4 3 ) 2 + … + C n n ( - 4 3 ) n = 2 C 0 n 7 n + C 2 n 7 n - 2 ( 4 3 ) 2 … … = 2 k , where k is an integer ∴ I + f + F is an even integer. ⇒ f + F is an integer since I is an integer. But 0 < f < 1 and 0 < F < 1 ⇒ f + F < 2 ∴ f + F = 1 ...(i) ⇒ I + 1 is an even integer. ∴ I is an odd integer. To prove, ( I + f ) ( 1 - f ) = 1 LHS = ( I + f ) ( 1 - f ) = ( 7 + 4 3 ) n ( 7 - 4 3 ) n = ( 7 + 4 3 ) ( 7 - 4 3 ) n = ( 49 - 48 ) n = 1 = RHS Hence, proved.

If n is a positive integer, prove that ∑ r = 1 n r 3 C r n C r - 1 n 2 = n ( n + 1 ) 2 ( n + 2 ) 12 .

To prove, ∑ r = 1 n r 3 C r n C r - 1 n 2 = n ( n + 1 ) 2 ( n + 2 ) 12 LHS = ∑ r = 1 n r 3 C r n C r - 1 n 2 = ∑ r = 1 n r 3 n - r + 1 r 2 = ∑ r = 1 n r ( n - r + 1 ) 2 = ∑ r = 1 n r ( n + 1 ) 2 - 2 ( n + 1 ) r + r 2 = ( n + 1 ) 2 Σ r - 2 ( n + 1 ) Σ r 2 + Σ r 3 = n + 1 2 ( n ) ( n + 1 ) 2 - 2 n + 1 ( n ) ( n + 1 ) ( 2 n + 1 ) 6 + n 2 ( n + 1 ) 2 4 = ( n + 1 ) 2 2 n n + 1 - 2 n ( 2 n + 1 ) 3 + n 2 2 = ( n + 1 ) 2 2 6 n 2 + 6 n - 8 n 2 - 4 n + 3 n 2 6 = ( n + 1 ) 2 2 n 2 + 2 n 6 = n ( n + 1 ) 2 ( n + 2 ) 12 = RHS Hence, proved.

E M B I B E

Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 6 - Binomial Theorem>Exercise 6(a)

Andhra Pradesh Board Solutions for Exercise 1: Exercise 6(a)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 1: Exercise 6(a)

Attempt the practice questions from Exercise 1: Exercise 6(a) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Exercise 1: Exercise 6(a) with Hints & Solutions

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12th Andhra Pradesh Board

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Intermediate Second Year Mathematics Paper 2A>Chapter 6 - Binomial Theorem>Exercise 6(a)>Q III

If ${(1 + 3 x - 2 x^{2})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ , then prove that $a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{20} = 4^{10}$ .