# what is the coefficient of x^2y^6 in (5x-2y)^8?

## Question:

What is the coefficient of {eq}x^2y^6 {/eq} in {eq}(5x-2y)^8 {/eq}?

## Binomial theorem:

Usually, we use normal multiplication to expand a binomial which is raised to an exponent. But if the exponent is a large number, then its very difficult to expand it using multiplication and binomial theorem to do this. The binomial theorem states:

{eq}(a+b)^{n}=n^{C} 0 a^{n}+n^{C} 1 a^{n-1} b+n^{C} 2 a^{n-2} b^{2}+n^{C} 3 a^{n-3} b^{3}+\ldots \ldots \ldots+n^{C} n b^{n} {/eq}, where

{eq}n^{C} r=\dfrac{n!}{(n-k)!k!} {/eq}

The given expression is,

{eq}(5x-2y)^8 = (a+b)^n {/eq}

The {eq}(r+1)^{th} {/eq} term of the expansion of {eq}(a+b)^n {/eq} using the binomial theorem is {eq}T_{r+1} = n^{C} r \, a^{n-r} \cdot b^r {/eq}.

Substitute {eq}a=5x, \, b=-2y {/eq} and {eq}n=8 {/eq} in the above equation:

\begin{align} T_{r+1} &= 8^{C} r\, (5x)^{8-r} \cdot (-2y)^r \\[0.5cm] &= 8^{C} r\, 5^{8-r} (-2)^r x^{8-r} y^r & \rightarrow (1) \end{align}

We want the coefficient of {eq}x^2 y^6 {/eq}.

So we set the exponents of the corresponding bases in the last two expressions equal:

$$8-r=2; \,\, r=6\\ r=6; \,\, r=6$$

So we got {eq}r=6 {/eq} from both the equations.

We substitute this in (1):

\begin{align} T_7 &= 8^{C} 6\,\, 5^{8-6}\, (-2)^6\, x^{8-6} \,y^6\\[0.5cm] &= \dfrac{8!}{(8-6)! \, 6!} \,5^2 \,2^6 \,x^2\, y^6 & [\because n^{C} r= \dfrac{n!}{(n-r)!r!} ] \\[0.5cm] &= \dfrac{8!}{2! \, 6!}\, (25)\,(64)\, (x^2y^6) \\[0.5cm] &= \dfrac{8 \times 7 \times 6!}{(2 \times 1)\, (6!)}\, (1600 x^2y^6) \\[0.5cm] &= 44800\, x^{2}\,y^{6} \end{align}

Therefore, the coefficient of {eq}x^{2} \,y^{6} {/eq} in the given expansion is: {eq}\color{blue}{\boxed{\mathbf{44,800 }}} {/eq}. 