Question

What is the 8th term of (a-b)^11??
please help :D
Tyy

Answer 1

`\large\text{$-330a^4b^7$}`

Explanation:

The expansion of (a - b)¹¹ can be found using the binomial theorem.

The general form of the binomial theorem is given by:

`\boxed{\begin{array}{c} \underline{\sf Binomial\; Theorem}\\\\\displaystyle (a+b)^n=\sum^(n)_(k=0)\binom{n}{k} a^(n-k)b^(k)\\\\\\\textsf{where}\;\displaystyle \binom{n}{k} = (n!)/(k!(n-k)!)\\\end{array}}`

In the case of (a - b)¹¹:

• 
  `a = a`
• 
  `b = -b`
• 
  `n = 11`

Substituting these values into the binomial theorem, we get:

`(a - b)^(11)=\displaystyle \sum^(11)_(k=0)\binom{11}{k} a^(11-k)(-b)^(k)`

As the first term corresponds to k = 0, then the 8th term corresponds to k = 7, so we can calculate the 8th term by substituting k = 7 into the equation:

`\begin{aligned}\displaystyle \binom{11}{7} a^(11-7) (-b)^7 &= \binom{11}{7} a^4 (-b)^7\\\\&=-(11!)/(7!(11-7)!)a^4b^7\\\\&=-(11!)/(7!\:4!)a^4b^7\\\\&=-(11*10*9*8*7*6*5*4*3*2*1)/(7*6*5*4*3*2*1*4*3*2*1)\:a^4b^7\\\\&=-(11*10*9*8*7*6*5)/(7*6*5*4*3*2*1)\:a^4b^7\\\\&=-(1663200)/(5040)a^4b^7\\\\&=-330a^4b^7\end{aligned}`

Therefore, the 8th term of (a - b)¹¹ is:

`\Large\boxed{\boxed{-330a^4b^7}}`

Answer 2

8th term:
`\sf - 330a^4b^7`

Explanation:

To find the 8th term of
`\sf (a-b)^(11)`, we can apply the binomial theorem, which states that:

`\sf (x+y)^n = \sum_(k=0)^(n) \binom{n}{k} \cdot x^(n-k) \cdot y^k`

In the given expression,
`\sf (a-b)^(11)`, we look for the 8th term, which corresponds to
`\sf k = 7`:

x = a and y = (-b).

Substitute the value:

`\sf \textsf{8th term} = \binom{11}{7} \cdot a^(11-7) \cdot (-b)^7`

Calculate the binomial coefficient:

`\sf \binom{11}{7} = (11!)/(7! \cdot (11-7)!) = 330`

Now, substitute back into the expression:

`\sf \textsf{8th term} = 330 \cdot a^4 \cdot -b^7`

`\sf \textsf{8th term} = -330 \cdot a^4 \cdot b^7`

Therefore, the 8th term of
`\sf (a-b)^(11)` is
`\sf - 330a^4b^7`.