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

Question

Abigail Allen · Accepted Answer

#(1.07)^7 ~~ 1.60578#

By the binomial theorem we have:

#(a+b)^7 = sum_(k=0)^7 ((7),(k)) 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:

#(a+b)^7 = a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7#

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

#1.07^7 ~~ 1+7(0.07)+21(0.07)^2+35(0.07)^3+35(0.07)^4+21(0.07)^5#

#=1+7(0.07)+21(0.0049)+35(0.000343)+35(0.00002401)+21(0.0000016807)#
#=1+0.49+0.1029+0.012005+0.00084035+0.0000352947#
#~~1.60578#

Eli Assouline · Answer

undefined To approximate $1.07^7$ using the binomial theorem, we use the formula:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

For $1.07^7$, $a = 1$ and $b = 0.07$. Applying the binomial theorem:

$$1.07^7 \approx \binom{7}{0} \cdot 1^{7-0} \cdot 0.07^0 + \binom{7}{1} \cdot 1^{7-1} \cdot 0.07^1$$

$$= 1 + 7 \cdot 1 \cdot 0.07 = 1 + 0.49 = 1.49$$