First slide

Binomial theorem for positive integral Index

Question

The value of (1.00 2)¹² up to fourth place of decimal is

Moderate

A

1.0242
B

1.0245
C

1.0004
D

1.0254

Solution

We have, (1.002)12 or it can be rewritten as (1 + 0.00212
$\begin{aligned} \Rightarrow (1 .002)^{12} = 1 +^{12} C_{1} (0 .002) +^{12} C_{2} (0 .002)^{2} +^{12} C_{3} (0 .002)^{3} + \dots . . . \end{aligned}$
we want the answer up to 4 decimal places and as such
we have left further expansion.
$\begin{aligned} ∴ (1 .002)^{12} = 1 + 12 (0 .002) + \frac{12 \cdot 11}{1 \cdot 2} (0 .002)^{2} + \frac{12 \cdot 11 \cdot 10}{1 \cdot 2 \cdot 3} (0 .002)^{3} + \dots \end{aligned}$
$\begin{aligned} = 1 + 0 .024 + 2 .64 \times 10^{- 4} + 1 .76 \times 10^{- 6} + \dots . . . = = 1 .0242 \end{aligned}$

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