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)

Telangana Board Solutions for Exercise 1: Exercise 6(a)

Author:Telangana Board

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

Attempt the free 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 Telangana Board Solutions for Exercise 1: Exercise 6(a) with Hints & Solutions

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12th Telangana 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 $a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{20} = 4^{10}$ .