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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 

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Solution

We notice that the coefficients 1, 4, 6, 4, 1 are the values of 4C0, 4C1, 4C2, 4C3, 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) 

= [(2x − 1) + (3 − 2x)]4

= (2x − 1 + 3 − 2x)4

= (2)4

= 16

Concept: Binomial Theorem for Positive Integral Index
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.2 | Q 7. (ii) | Page 77
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