# How do you use the binomial theorem to approximate the value of 1.07^7 ?

May 9, 2016

${\left(1.07\right)}^{7} \approx 1.60578$

#### Explanation:

By the binomial theorem we have:

${\left(a + b\right)}^{7} = {\sum}_{k = 0}^{7} \left(\begin{matrix}7 \\ k\end{matrix}\right) {a}^{7 - k} {b}^{k}$

where ((7),(k)) = (7!)/(k!(7-k)!)

We can get these binomial coefficients from the row of Pascal's triangle that begins $1 , 7$. Some people call this the $7$th row (calling the first row the $0$th). Personally I prefer to call it the $8$th row, but regardless of what you call it, it's the one that begins $1 , 7$:

So:

${\left(a + b\right)}^{7} = {a}^{7} + 7 {a}^{6} b + 21 {a}^{5} {b}^{2} + 35 {a}^{4} {b}^{3} + 35 {a}^{3} {b}^{4} + 21 {a}^{2} {b}^{5} + 7 a {b}^{6} + {b}^{7}$

Putting $a = 1$ and $b = 0.07$ we have:

${1.07}^{7} \approx 1 + 7 \left(0.07\right) + 21 {\left(0.07\right)}^{2} + 35 {\left(0.07\right)}^{3} + 35 {\left(0.07\right)}^{4} + 21 {\left(0.07\right)}^{5}$

$= 1 + 7 \left(0.07\right) + 21 \left(0.0049\right) + 35 \left(0.000343\right) + 35 \left(0.00002401\right) + 21 \left(0.0000016807\right)$

$= 1 + 0.49 + 0.1029 + 0.012005 + 0.00084035 + 0.0000352947$

$\approx 1.60578$