Evaluate x + x 2 - 1 6 + x - x 2 - 1 6 .

Given, x + x 2 - 1 6 + x - x 2 - 1 6 = ∑ r = 0 6 C r 6 x 6 - r . x 2 - 1 r + ∑ r = 0 6 C r 6 x 6 - r . - x 2 - 1 r = C 0 6 x 6 + C 1 6 x 5 x 2 - 1 ) + C 2 6 x 4 ( x 2 - 1 ) 2 + - - - - + C 6 6 x 2 - 1 6 + C 0 6 x 6 - C 1 6 x 5 x 2 - 1 ) + C 2 6 x 4 ( x 2 - 1 ) 2 + - - - - - + C 6 6 x 2 - 1 6 = 2 C 0 6 x 6 + C 2 6 x 4 x 2 - 1 + C 4 6 x 2 x 2 - 1 2 + C 6 6 x 2 - 1 3 = 2 x 6 + 156 x 6 - 15 x 4 + 15 x 6 + 15 x 2 - 30 x 4 + x 6 - 1 - 3 x 4 + 3 x 2 = 2 32 x 6 - 48 x 4 + 18 x 2 - 1 = 64 x 6 - 96 x 4 + 36 x 2 - 2

Evaluate 1 + 2 x 5 + 1 - 2 x 5 .

Given, ( 1 + 2 x ) 5 + ( 1 - 2 x ) 5 = C 0 5 + C 1 5 2 x + C 2 5 2 x 2 + - - - + C 5 5 2 x 5 + C 0 5 - C 1 5 2 x + C 2 5 2 x 2 + - - - C 5 5 2 x 5 = 2 C 0 5 + C 2 5 2 x 2 + C 4 5 2 x 4 = 2 1 + 10 × 4 × x + 5 × 16 x 2 = 2 1 + 40 x + 80 x 2 = 160 x 2 + 80 x + 2

Evaluate a 2 + a 2 - 1 4 + a 2 - a 2 - 1 4 .

Given, a 2 + a 2 - 1 4 + a 2 - a 2 - 1 4 Let a 2 = x and a 2 - 1 = y ⇒ a 2 + a 2 - 1 4 + a 2 - a 2 - 1 4 = x + y 4 + x - y 4 = C 0 4 x 4 + C 1 4 x 3 y + C 2 4 x 2 y 2 + - - - + C 4 4 y 4 + C 0 4 x 4 - C 1 4 x 3 y + C 2 4 x 2 y 2 + - - - + C 4 4 y 4 = 2 C 0 4 x 4 y 0 + C 2 4 x 2 y 2 + C 4 4 x 0 y 4 = 2 x 4 y 0 + 6 x 2 y 2 + x 0 y 4 = 2 a 8 + 6 a 4 a 2 - 1 + a 2 - 1 2 = 2 a 8 + 6 a 6 - 6 a 4 + a 4 + 1 - 2 a 2 = 2 a 8 + 6 a 6 - 5 a 4 - 2 a 2 + 1 = 2 a 8 + 12 a 6 - 10 a 4 - 4 a 2 + 2

E M B I B E

MATHEMATICS for CLASS XI>Chapter 18 - Binomial Theorem>EXERCISE 18.1

R D Sharma Solutions for Chapter: Binomial Theorem, Exercise 1: EXERCISE 18.1

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Binomial Theorem, Exercise 1: EXERCISE 18.1

Attempt the practice questions on Chapter 18: Binomial Theorem, Exercise 1: EXERCISE 18.1 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.