Maharashtra State BoardHSC Arts 11th
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Find the value of (1.02)6, correct upto four places of decimal - Mathematics and Statistics

Sum

Find the value of (1.02)6, correct upto four places of decimal

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Solution

(1.02)6 = (1 + 0.02)6

= 6C0(1)6(0.02)0 + 6C1(1)5(0.02)1 + 6C2(1)4(0.02)2 + 6C3(1)3(0.02)3 + 6C4(1)2(0.02)4 + 6C5(1)1(0.02)5 + 6C6(1)0(0.02) 

Since, 6C0 = 6C6 = 1, 6C1 = 6C5 = 6,

6C2 = 6C4 = `(6 xx 5)/(2 xx 1)` = 15, 6C3 = `(6 xx 5 xx 4)/(3 xx 2 xx 1)` = 20

∴ (1.02)6 = 1(1)(1) + 6(1)(0.02) + 15(1)(0.0004) + 20(1)(0.000008) + ……

= 1 + 0.12 + 0.0060 + 0.000160 + ……

= 1.12616

= 1.1262, correct upto 4 decimal places.

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 8 | Page 77
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