Without expanding, find the value of (2x − 1)4 + 4(2x − 1)3 (3 − 2x) + 6(2x − 1)2 (3 − 2x)2 + 4(2x − 1)1 (3 − 2x)3 + (3 − 2x)4 - Mathematics and Statistics

Sum

Without expanding, find the value of (2x − 1)⁴ + 4(2x − 1)³ (3 − 2x) + 6(2x − 1)² (3 − 2x)² + 4(2x − 1)¹ (3 − 2x)³ + (3 − 2x)⁴

We notice that the coefficients 1, 4, 6, 4, 1 are the values of ⁴C₀, ⁴C₁, ⁴C₂, ⁴C_3, and ⁴C₄ respectively.

Hence, the given expression can be written as:

⁴C₀(2x − 1)⁴ + ⁴C₁(2x − 1)³(3 − 2x) + ⁴C₂(2x − 1)²(3 − 2x)² + ⁴C₃(2x − 1)(3 − 2x)³ + ⁴C₄(3 − 2x)⁴

= [(2x − 1) + (3 − 2x)]⁴

= (2x − 1 + 3 − 2x)⁴

= (2)⁴

= 16

Concept: Binomial Theorem for Positive Integral Index

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Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.2 | Q 7. (ii) | Page 77