Chapter 9.5, Problem 34E

To determine

To expand: The expression ${\left(x^2+y^2\right)}^5$ by using Pascal’s triangle to determine the coefficients.

Answer to Problem 34E

The expansion of the expression ${\left(x^2+y^2\right)}^5$ is $x^{10}+5x^8{y}^2+10x^6{y}^4+10x^4{y}^6+5x^2{y}^8+y^{10}$ .

Explanation of Solution

Given information: The expression is ${\left(x^2+y^2\right)}^5$ .

Formula used:Binomial Theorem: ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\cdots +nx{y}^{n-1}+y^n$ .

Calculation:

Here, $n=5$ .

First write out the Pascal’s triangle such that the row that begins with 1, 5.

  $\begin{array}{ccccccccccc}& & & & & 1& & & & & \\ & & & & 1& & 1& & & & \\ & & & 1& & 2& & 1& & & \\ & & 1& & 3& & 3& & 1& & \\ & 1& & 4& & 6& & 4& & 1& \\ 1& & 5& & 10& & 10& & 5& & 1\end{array}$

From the Pascal’s triangle, it is observed that the coefficients of the expression are 1, 5, 10, 10, 5 and 1.

Expand the expression ${\left(x^2+y^2\right)}^5$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(x^2+y^2\right)}^5=\left(1\right){\left(x^2\right)}^5+5{\left(x^2\right)}^4\left(y^2\right)+10{\left(x^2\right)}^3{\left(y^2\right)}^2+10{\left(x^2\right)}^2{\left(y^2\right)}^3+5{\left(x^2\right)}^1{\left(y^2\right)}^4+1{\left(y^2\right)}^5\\ =x^{10}+5x^8{y}^2+10x^6{y}^4+10x^4{y}^6+5x^2{y}^8+y^{10}\end{array}$

Thus, the expansion of the expression ${\left(x^2+y^2\right)}^5$ is $x^{10}+5x^8{y}^2+10x^6{y}^4+10x^4{y}^6+5x^2{y}^8+y^{10}$ .