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Use Pascal's Triangle to expand
(3+y^(2))^(5). Express your answer in simplest form.
Answer:

Use Pascal's Triangle to expand $\left(3+y^{2}\right)^{5}$ . Express your answer in simplest form. $\newline$ Answer:

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Pascal's triangle and the Binomial Theorem

Full solution

Q. Use Pascal's Triangle to expand

\left(3+y^{2}\right)^{5}

. Express your answer in simplest form.

\newline

Answer:

Identify Coefficients: Identify the coefficients from Pascal's Triangle for the expansion of $(a+b)^5$ . The coefficients for the expansion of a binomial to the fifth power are $1, 5, 10, 10, 5, 1$ .
Write General Form: Write down the general form of the binomial expansion using the coefficients. $\newline$ The general form of the expansion is: $\newline$ $(3+y^2)^5 = 1\cdot(3^5)\cdot(y^2)^0 + 5\cdot(3^4)\cdot(y^2)^1 + 10\cdot(3^3)\cdot(y^2)^2 + 10\cdot(3^2)\cdot(y^2)^3 + 5\cdot(3^1)\cdot(y^2)^4 + 1\cdot(3^0)\cdot(y^2)^5$
Calculate Each Term: Calculate each term of the expansion. $\newline$ $1$ st term: $1*(3^5)*(y^2)^0 = 1*243*1 = 243$ $\newline$ $2$ nd term: $5*(3^4)*(y^2)^1 = 5*81*y^2 = 405y^2$ $\newline$ $3$ rd term: $10*(3^3)*(y^2)^2 = 10*27*y^4 = 270y^4$ $\newline$ $4$ th term: $10*(3^2)*(y^2)^3 = 10*9*y^6 = 90y^6$ $\newline$ $5$ th term: $5*(3^1)*(y^2)^4 = 5*3*y^8 = 15y^8$ $\newline$ $6$ th term: $1*(3^0)*(y^2)^5 = 1*1*y^{10} = y^{10}$
Combine Terms: Combine all the terms to write the final expanded form. $\newline$ $(3+y^2)^5 = 243 + 405y^2 + 270y^4 + 90y^6 + 15y^8 + y^{10}$

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