Question

Without simplifying, find:
a) the 6th term of (2x+5)^15
b) the 4th term of (x^2+5/x)^9

Answer 1

Part A)

`\displaystyle z_6 = 9609600000x^(10)`

Part B)

`z_4 = 10500 x^9`

Explanation:

Recall the binomial expansion theorem:

`\displaystyle (x+y)^n = \sum_(k=0)^(n){n \choose k} x^(n-k) y^k`

Part A)

Our expression is equivalent to:

`\displaystyle (2x+5)^(15) = \sum_(k = 0)^(15) {15 \choose k} (2x)^(15-k)\cdot 5^k`

To find the sixth term, let k = 5. Therefore, the sixth term is:

`\displaystyle \begin{aligned} z_6 &= {15\choose 5} (2x)^(15-5)\cdot 5^5 \\ \\ & = {15\choose 5}x^(10) \cdot (2)^(10)\cdot 5^5 \\ \\ &= 9609600000x^(10)\end{aligned}`

Part B)

Likewise:

`\displaystyle \begin{aligned} \left(x^2 + (5)/(x)\right)^9 = \sum_(k=0)^9 {9\choose k}(x^2)^(9-k)\left((5)/(x)\right)^(k)\end{aligned}`

To find the fourth term, let k = 3. Therefore, the fourth term is:

`\displaystyle \begin{aligned} z_4 & = {9\choose 3}\left(x^2\right)^(9-3) \left((5)/(x)\right)^(3) \\ \\ & = {9\choose 3}x^(12) \cdot (5^3)/(x^3) \\ \\ & = 10500x^9\end{aligned}`