We have to expand the expression using the Binomial Theorem. So use the binomial formula,

`(a+b)^n=sum_(k=0)^n((n),(k))a^(n-k)b^k`

`:.(x+1)^6=((6),(0))x^(6-0)*1^0+((6),(1))x^(6-1)*1^(1)+((6),(2))x^(6-2)*1^2+((6),(3))x^(6-3)*1^3+((6),(4))x^(6-4)*1^4+((6),(5))x^(6-5)*1^5+((6),(6))x^(6-6)*1^6`

`=x^6+(6!)/(1!(6-1)!)x^5+(6!)/(2!(6-2)!)x^4+(6!)/(3!(6-3)!)x^3+(6!)/(4!(6-4)!)x^2+(6!)/(5!(6-5)!)x^1+1`

`=x^6+(6*5!)/(5!)x^5+(6*5*4!)/(2*1*4!)x^4+(6*5*4*3!)/(3*2*1*3!)x^3+(6*5*4!)/(4!*2*1)x^2+(6*5!)/(5!)x+1`

`=x^6+6x^5+15x^4+20x^3+15x^2+6x+1`

## See

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We have to expand the expression using the Binomial Theorem. So use the binomial formula,

`(a+b)^n=sum_(k=0)^n((n),(k))a^(n-k)b^k`

`:.(x+1)^6=((6),(0))x^(6-0)*1^0+((6),(1))x^(6-1)*1^(1)+((6),(2))x^(6-2)*1^2+((6),(3))x^(6-3)*1^3+((6),(4))x^(6-4)*1^4+((6),(5))x^(6-5)*1^5+((6),(6))x^(6-6)*1^6`

`=x^6+(6!)/(1!(6-1)!)x^5+(6!)/(2!(6-2)!)x^4+(6!)/(3!(6-3)!)x^3+(6!)/(4!(6-4)!)x^2+(6!)/(5!(6-5)!)x^1+1`

`=x^6+(6*5!)/(5!)x^5+(6*5*4!)/(2*1*4!)x^4+(6*5*4*3!)/(3*2*1*3!)x^3+(6*5*4!)/(4!*2*1)x^2+(6*5!)/(5!)x+1`

`=x^6+6x^5+15x^4+20x^3+15x^2+6x+1`

A simple method used to solve this binomial is the use of pascal's triangle:

The expansion of our example is expanded as follows using pascal's triangle:

`(a + b)^6 = 1a^6 + 6a^5b^1 + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6a^1b^5 + 1b^6`

Now let's use the above expansion to solve our problem:

Once we evaluate above, we obtain our final answer as:

`(x+1)^6=x^6+6x^5+15x^4+20x^3+15x^2+6x+1`

Additionally, there is an equation that can be used to expand a binomial theorem. This equation is expressed as a formula:

where:

a= first term

b= last term

n = exponent (power in original equation)

k = term required - 1

At times this equation can be more tedious than using pascal's triangle, however once 'n' get's bigger and bigger it becomes more difficult to use pascal's triangle and the above equation is used.