Question

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

Answer 1

according to Binomial expansion,

(1 + x)ⁿ = 1 + nx + n(n - 1)x²/2! + n(n - 1)(n - 2)x³/3! + n(n - 1)(n - 2)(n - 3)x⁴/4! + .........∞

here, (1.02)⁶ = (1 + 0.02)⁶

x = 0.02 , n = 6

so, (1 + 0.02)⁶ = 1 + 6(0.02) + 6(6 - 1)(0.02)²/2! + 6(6 - 1)(6 - 2)(0.02)³/3! + ....

= 1 + 0.12 + 6 × 5 (0.0004)/2! + 6 × 5 × 4 × (0.000008)/3! + .....

= 1 + 0.12 + 30 × 0.0004/2 + 120 × 0.000008/6 + ... .

[as we know, 2! = 2, 3! = 6, 4! = 24 .... ]

= 1 + 0.12 + 0.0060 + 0.00016 + ...

= 1.12616.....

but we have to find (1.02)⁶ upto four place of decimals .

so, value of (1.02)⁶ = 1.1261

Answer 2

Answer:-

Value of (1.02)⁶ = 1.1261

Step-by-step explanation:

In this question

We have to find the value of (1.02)⁶ up to four decimal places.

Using Binomial expansion,

(1 + x)ⁿ = $1\ +\ nx\ +\ \frac{n(n - 1)x^{2}}{2!}\ +\ \frac{n(n - 1)(n - 2)x^{3}}{3!}\ +\ \frac{n(n - 1)(n - 2)(n - 3)x^{4}}{4!}\ + .........$

Here, (1.02)⁶ = (1 + 0.02)⁶

x = 0.02 , n = 6

So, (1 + 0.02)⁶ = $1\ +\ 6(0.02)\ +\ \frac{6(6 - 1)(0.02)^{2}}{2!} +\ \frac{6(6 - 1)(6 - 2)(0.02)^{3}}{3! }+ ....$

= $1\ +\ 0.12\ +\ \frac{6\times 5 (0.0004)}{2!}\ +\ \frac{6\times 5\times 4\times (0.000008)}{3!}+ .....$

= $1\ +\ 0.12\ +\ \frac{30\times 0.0004}{2}\ +\ \frac{120\times 0.000008}{6} + ....$

= 1 + 0.12 + 0.0060 + 0.00016 + ...

= 1.12616.....

But we have to find (1.02)⁶ up to four place of decimals .

So, value of (1.02)⁶ = 1.1261