Question

Find the 7th term in the following binomial expansion(4f+t/3)8 in ascending powers of t

Answer 1

Answer:

$\displaystyle T_6=\frac{448}{729} f^2t^6$

Step-by-step explanation:

Binomial Theorem

Any power of x + y can be expanded into a sum of the form:

$\displaystyle (x+y)^{n}={n \choose 0}x^{n}y^{0}+{n \choose 1}x^{n-1}y^{1}+{n \choose 2}x^{n-2}y^{2}+\cdots +{n \choose n-1}x^{1}y^{n-1}+{n \choose n}x^{0}y^{n}$

For any given term k, counted from 0 to n:

$\displaystyle T_k={n \choose k}x^{n-k}y^{k}$

Note if we have to find the 7th term, we have to use k=6. The expression is:

$\displaystyle \left(4f+\frac{t}{3}\right)^8$

For n=8, x=4f, y=t/3, k=6:

$\displaystyle T_6={8 \choose 6}(4f)^{8-6}\left (\frac{t}{3} \right )^{6}$

$\displaystyle T_6={8 \choose 6}(4f)^{2}\left (\frac{t}{3} \right )^{6}$

$\displaystyle T_6={8 \choose 6}16f^2\frac{t^6}{3^6}$

Calculate:

$\displaystyle {8 \choose 6}=\frac{8!}{2!\cdot 6!}=28$

Thus:

$\displaystyle T_6=28\cdot 16f^2\frac{t^6}{729}$

Simplifying, the 7th term is:

$\mathbf{\displaystyle T_6=\frac{448}{729} f^2t^6}$