Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Use Pascal's Triangle to expand 
(4y^(2)+1)^(3). Express your answer in simplest form.
Answer:

Use Pascal's Triangle to expand (4y2+1)3 \left(4 y^{2}+1\right)^{3} . Express your answer in simplest form.\newlineAnswer:

Full solution

Q. Use Pascal's Triangle to expand (4y2+1)3 \left(4 y^{2}+1\right)^{3} . Express your answer in simplest form.\newlineAnswer:
  1. Identify Row: Identify the row of Pascal's Triangle that corresponds to the exponent of the binomial expansion.\newlineSince we are expanding (4y2+1)3(4y^{2}+1)^{3}, we need the 4th4^{\text{th}} row of Pascal's Triangle, which corresponds to the coefficients for a cubic expansion.\newlineThe 4th4^{\text{th}} row of Pascal's Triangle is 1,3,3,11, 3, 3, 1.
  2. Write Expansion Terms: Write out each term of the expansion using the binomial theorem and the coefficients from Pascal's Triangle.\newlineThe binomial theorem states that (a+b)n=Σ(nk)ankbk(a+b)^n = \Sigma \binom{n}{k} \cdot a^{n-k} \cdot b^k, where Σ\Sigma denotes the sum over kk from 00 to nn.\newlineUsing the coefficients from Pascal's Triangle, the expansion will be:\newline1(4y2)3(1)0+3(4y2)2(1)1+3(4y2)1(1)2+1(4y2)0(1)31\cdot(4y^2)^3\cdot(1)^0 + 3\cdot(4y^2)^2\cdot(1)^1 + 3\cdot(4y^2)^1\cdot(1)^2 + 1\cdot(4y^2)^0\cdot(1)^3
  3. Calculate Each Term: Calculate each term of the expansion.\newline1×(4y2)3×(1)0=1×(64y6)×(1)=64y61\times(4y^2)^3\times(1)^0 = 1\times(64y^6)\times(1) = 64y^6\newline3×(4y2)2×(1)1=3×(16y4)×(1)=48y43\times(4y^2)^2\times(1)^1 = 3\times(16y^4)\times(1) = 48y^4\newline3×(4y2)1×(1)2=3×(4y2)×(1)=12y23\times(4y^2)^1\times(1)^2 = 3\times(4y^2)\times(1) = 12y^2\newline1×(4y2)0×(1)3=1×(1)×(1)=11\times(4y^2)^0\times(1)^3 = 1\times(1)\times(1) = 1
  4. Combine Terms for Final Form: Combine all the terms to write the final expanded form.\newlineThe expanded form of (4y2+1)3(4y^{2}+1)^{3} is:\newline64y6+48y4+12y2+164y^6 + 48y^4 + 12y^2 + 1

More problems from Pascal's triangle and the Binomial Theorem