Problem 30 - Algebra and Trigonometry

Question

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (a+2 b)^{6} $$

Step-by-Step Solution

Answer

(a+2b)^6 equals to a^6 + 12a^5b + 60a^4b^2 + 160a^3b^3 + 240a^2b^4 + 192a*b^5 + 64b^6

Step 1: Identify p, q and n

p corresponds to the term 'a', q corresponds to the term '2b', and n corresponds to '6'.

Step 2: Use the Binomial Theorem

(a+2b)^6 = Σ (from k=0 to 6) [6Ck * a^(6-k) * (2b)^k]. This is based on the Binomial Theorem.

Step 3: Expand the expression

Start off with k=0 and increment k by 1 each step until reaching 6. Calculate each term independently, then add these terms together to get the simplified form of the expansion.

Step 4: Solve for each term

Term for k=0: 6C0 * a^(6-0) * (2b)^0= 1*a^6*1=a^6.\nTerm for k=1: 6C1 * a^(6-1) * (2b)^1 = 6*a^5*(2b) = 12a^5b.\nTerm for k=2: 6C2*a^(6-2)*(2b)^2 = 15*a^4*(2*2)b^2 = 60a^4b^2.\nTerm for k=3: 6C3*a^(6-3)*(2b)^3 = 20*a^3*(2*2*2)b^3 = 160a^3b^3. \nTerm for k=4: 6C4*a^(6-4)*(2b)^4 = 15*a^2*(2*2*2*2)b^4 = 240a^2b^4. \nTerm for k=5: 6C5*a^(6-5)*(2b)^5 = 6*a*(2*2*2*2*2)b^5 = 192a*b^5. \nTerm for k=6: 6C6*a^(6-6)*(2b)^6 = 1*(2*2*2*2*2*2)b^6 = 64b^6.

Step 5: Add all terms

Add all these terms together to get the final expanded form: (a+2b)^6 = a^6 + 12a^5b + 60a^4b^2 + 160a^3b^3 + 240a^2b^4 + 192a*b^5 + 64b^6