# Specific Term of a binomial expansion:

#### Finding a particular term of the binomial expansion $$(a + b)^n$$ from starting point:

Let us have to find out the "$$k^{th}$$" term of the binomial expansion then. $$T_k = \binom{n}{k - 1} \ a^{(n - k + 1)} \ b^{(k - 1)}$$

Note: The total number of terms in the binomial expansion $$(a + b)^n$$ will always be $$(n + 1)$$.

Example(1): Find the fourth term of the binomial $$(4x + 5)^5 ?$$

Solution: If a question does not say to find any particular term from the starting point or from the end of the binomial expansion then we can simply find that particular term from the starting point.

Given $$n = 5$$, $$k = 4$$, $$a = 4x$$, and $$b = 5$$.

By putting these values into the formula. $$T_k = \binom{n}{k - 1} \ a^{(n - k + 1)} \ b^{(k - 1)}$$ $$T_4 = \binom{5}{4 - 1} \ (4x)^{(5 - 4 + 1)} \ 5^{(4 - 1)}$$ $$T_4 = \binom{5}{3} \ (4x)^2 \ 5^3$$ $$T_4 = \frac{5!}{(5 - 3) \ 3!} \times 16 x^2 \times 125$$ $$T_4 = \frac{5!}{2! \ 3!} \times 16 x^2 \times 125$$ $$T_4 = \frac{5 \times 4 \times 3!}{2! \ 3!} \times 16 x^2 \times 125$$ $$T_4 = \frac{5 \times 4}{2 \times 1} \times 16 x^2 \times 125$$ $$T_4 = \frac{20}{2} \times 16 x^2 \times 125$$ $$T_4 = 20,000 x^2$$ Here the total number of terms is six, and hence $$20,000 x^2$$ is the fourth term from the starting point.

Example(2): Find the sixth term of the binomial $$(x + 2y)^8 ?$$

Solution: Given $$n = 8$$, $$k = 6$$, $$a = x$$, and $$b = 2y$$.

By putting these values into the formula. $$T_k = \binom{n}{k - 1} \ a^{(n - k + 1)} \ b^{(k - 1)}$$ $$T_6 = \binom{8}{6 - 1} \ (x)^{(8 - 6 + 1)} \ (2y)^{(6 - 1)}$$ $$T_6 = \binom{8}{5} \ x^3 \ (2y)^5$$ $$T_6 = \frac{8!}{(8 - 5) \ 5!} \times x^3 \times (2y)^5$$ $$T_6 = \frac{8!}{3! \ 5!} \times x^3 \times 32 y^5$$ $$T_6 = \frac{8 \times 7 \times 6 \times 5!}{3! \ 5!} \times x^3 \times 32 y^5$$ $$T_6 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times x^3 \times 32 y^5$$ $$T_6 = 1792 x^3 y^5$$ Here the total number of terms is nine, and hence $$1792 x^3 y^5$$ is the sixth term from the starting point.

#### Finding a particular term of the binomial expansion $$(a + b)^n$$ from the end:

Let us have to find out the "$$k^{th}$$" term of the binomial expansion from the end then.

$$k^{th}$$ term from the end of the binomial expansion = $$(n - k + 2)^{th}$$ term from the starting point of the expansion.

Note: The total number of terms in the binomial expansion $$(a + b)^n$$ will always be $$(n + 1)$$.

Example(1): Find the fourth term from the last of the binomial $$(x + y)^6 ?$$

Solution: Given $$n = 6$$, $$k = 4$$, $$a = x$$, and $$b = y$$.

$$4^{th}$$ term from the last = $$(6 - 4 + 2)^{th}$$ term from the starting of the binomial expansion.

$$4^{th}$$ term from the last = $$4^{th}$$ term from the starting of the binomial expansion.

Now by putting the values into the formula of $$T_k$$. $$T_k = \binom{n}{k - 1} \ a^{(n - k + 1)} \ b^{(k - 1)}$$ $$T_4 = \binom{6}{4 - 1} \ (x)^{(6 - 4 + 1)} \ (y)^{(4 - 1)}$$ $$T_4 = \binom{6}{3} \ x^3 \ (y)^3$$ $$T_4 = \frac{6!}{(6 - 3) \ 3!} \times x^3 \times (y)^3$$ $$T_4 = \frac{6!}{3! \ 3!} \times x^3 \times y^3$$ $$T_4 = \frac{6 \times 5 \times 4 \times 3!}{3! \ 3!} \times x^3 \times y^3$$ $$T_4 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times x^3 \times y^3$$ $$T_4 = 20 x^3 y^3$$ Here the total number of terms is seven, and hence $$20 x^3 y^3$$ is the fourth term from the end.