Use binomial theorem to evaluate the following upto four places of decimal (0.98)–3

Use binomial theorem to evaluate the following upto four places of decimal

(0.98)^–3

Sum

(0.98)^–3

= (1 – 0.02)^–3

= `1 - (-3) (0.02) + ((-3)(-3 - 1))/(2!) (0.02)^2 - ((-3)(-3 - 1)(-3 - 2))/(3!) (0.02)^3 + ......`

= `1 + 0.06 + (-3(-4))/2 (0.02)^2 - ((-3)(-4)(-5))/6 (0.02)^3 + ......`

= 1 + 0.06 + 0.0024 + 0.00008 + ….

= 1.06248

= 1.0625

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Binomial Theorem for Negative Index Or Fraction

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Chapter 4: Methods of Induction and Binomial Theorem - Exercise 4.4 [Page 82]

Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.4 | Q 4. (v) | Page 82

State, by writing first four terms, the expansion of the following, where |x| < 1

(1 + x)⁻⁴

State, by writing first four terms, the expansion of the following, where |x| < 1

(1 + x²)^–1

State, by writing first four terms, the expansion of the following, where |b| < |a|

(a − b)⁻³

State, by writing first four terms, the expansion of the following, where |b| < |a|

(a + b)⁻⁴

State, by writing first four terms, the expansion of the following, where |b| < |a|

`("a" + "b")^(1/4)`

State, by writing first four terms, the expansion of the following, where |b| < |a|

`("a" - "b")^(-1/4)`