Maharashtra State BoardHSC Science (Electronics) 11th
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Find the value of (1.01)5, correct up to three places of decimals. - Mathematics and Statistics

Sum

Find the value of (1.01)5, correct up to three places of decimals.

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Solution

(1.01)5 = (1 + 0.01)5

= 5C0(1)5 + 5C1(1)4(0.01) + 5C2(1)3(0.01)2 + 5C3(1)2(0.01)3 + 5C4(1)(0.01)4 + 5C5(0.01)5

Now, 5C0 = 1 = 5C5

5C1 = 5 = 5C4

5C2 = `(5 xx 4)/(1 xx 2)` = 10 = 5C3

∴ (1.01)5 = 1 × 1 + 5 × 1 × 0.01 + 10 × 1 × 0.0001 + 10 × 1 × 0.000001 + ...

= 1 + 0.05 + 0.001 + 0.00001 + ...

= 1.05101...

= 1.051, correct up to three places of decimals.

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 9 | Page 77
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