Using the binomial theroem to estimate 6.07^2 ive got 6^2(1+0.01)^2 but im not sure if its right and i don't know where to go for here

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Using the binomial theroem to estimate 6.07^2

ive got 6^2(1+0.01)^2
but im not sure if its right and i don't know where to go for here

Reiny · Answer

You will have to know the special case of the general binomial expansion, namely

(1 + x)^n = 1 + nx + n(n-1)/2!x^2 + n(n-1)(n-2)/3!x^3 + ..

if │x│ < 1 , then the terms become rapidly smaller and less significant as you continue down the series

so 6.07^2
= 6^2(1 + .011667)^2
= 36[1 + 2(.011667 + .011667^2) + 0 + 0 + ...]

(notice if n=2, then one of the factors in n(n-1)(n-2).. becomes zero and all subsequent terms are zero)

so now evaluate 36(1 + 2(.011667)
to get 36.84

PS:
This is a very silly question. The calculations needed for your estimation are more difficult than those needed to just multiply 6.07 x 6.07

The binomial theorem was an important tool in finding roots and huge powers when we had no calculators, but to use it in such a question as the one you posted makes very little sense

Using the binomial theroem to estimate 6.07^2

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