# In the expansion of (ax+by)^7, the coefficients of the first 2 terms are 127 and -224, respectively. Find values of a and b?

Mar 27, 2018

a = root(7)(127) = 127^(1/7)~~1.998;
$b = - \frac{32}{127} ^ \left(\frac{6}{7}\right) = - 32 \frac{\sqrt{127}}{127} \approx - 0.503$

#### Explanation:

Given: binomial expansion: ${\left(a x + b y\right)}^{7}$ ; 1st coefficient $= 127$;
2nd coefficient $= - 224$

Binomial expansion for ${\left(a x + b y\right)}^{7} :$

${=}_{7} {C}_{0} {\left(a x\right)}^{7} {\left(b y\right)}^{0} {+}_{7} {C}_{1} {\left(a x\right)}^{6} {\left(b y\right)}^{1} {+}_{7} {C}_{2} {\left(a x\right)}^{5} {\left(b y\right)}^{2} + \ldots {+}_{7} {C}_{7} {\left(a x\right)}^{0} {\left(b y\right)}^{7}$

${=}_{7} {C}_{0} {a}^{7} {x}^{7} {+}_{7} {C}_{1} {a}^{6} {x}^{6} b y {+}_{7} {C}_{2} {a}^{5} {x}^{5} {b}^{2} {y}^{2} + \ldots {+}_{7} {C}_{7} {b}^{7} {y}^{7}$

Combinations:
 " "_7C_0 = (7!)/((7-0)!0!) = (7!)/(7!) = 1 = _7C_7

" "_7C_1 = (7!)/((7-1)!1!) = (7!)/(6!) = (7*6!)/(6!) = 7

" "_7C_2 = (7!)/((7-2)!2!) = (7!)/(5!2*1) = (7*6*5!)/(5!*2) = 21

Binomial expansion for ${\left(a x + b y\right)}^{7} :$

$= {a}^{7} {x}^{7} + 7 {a}^{6} b {x}^{6} y + 21 {a}^{5} {b}^{2} {x}^{5} {y}^{2} + \ldots + {b}^{7} {y}^{7}$

Using the given coefficients for the first two terms, find $a \text{ & } b$:

a^7 = 127; " " 7 a^6b = -224

$a = \sqrt{127} = {\left(127\right)}^{\frac{1}{7}}$

$b = - \frac{224}{7 {a}^{6}} = - \frac{224}{7 {\left(\sqrt{127}\right)}^{6}} = - \frac{32}{127} ^ \left(\frac{6}{7}\right) \cdot \frac{\sqrt{127}}{\sqrt{127}}$
$b = \frac{- 32 \sqrt{127}}{127}$