Question

Expand (2x+2)^6
How would you find the answer using the binomial theorem?

Answer 1

Explanation:

`\displaystyle\\\sum\limits _(k=0)^n(n!)/(k!*(n-k)!)a^(n-k)b^k .\\\\k=0\\(n!)/(0!*(n-0)!)a^(n-0)b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\(n!)/(1!*(n-1)!) a^(n-1)b^1=C_n^1a^(n-1)b^1.\\\\k=2\\(n!)/(2!*(n-2)!) a^(n-2)b^2=C_n^2a^(n-2)b^2.\\\\k=n\\(n!)/(n!*(n-n)!) a^(n-n)b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^(n-1)b^1+C_n^2a^(n-2)b^2+...+C_n^nb^n=(a+b)^n.`

`\displaystyle\\(2x+2)^6=(6!)/((6-0)!*0!) (2x)^62^0+(6!)/((6-1)!*1!) (2x)^(6-1)2^1+(6!)/((6-2)!*2!)(2x)^(6-2)2^2+\\\\ +(6!)/((6-3)!*3!) (2a)^(6-3)2^3+(6!)/((6-4)*4!) (2x)^(6-4)b^4+(6!)/((6-5)!*5!)(2x)^(6-5) b^5+(6!)/((6-6)!*6!)(2x)^(6-6)b^6. \\\\`

`(2x+2)^6=(6!)/(6!*1) 2^6*x^6*1+(5!*6)/(5!*1)2^5*x^5*2+\\\\+(4!*5*6)/(4!*1*2)2^4*x^4*2^2+ (3!*4*5*6)/(3!*1*2*3) 2^3*x^3*2^3+(4!*5*6)/(2!*4!)2^2*x^2*2^4+\\\\+(5!*6)/(1!*5!) 2^1*x^1*2^5+(6!)/(0!*6!) x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.`