**Find the coefficient of x**^{5}in (x + 3)^{8}

Solution:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n }is given by T_{r+1} = ^{n}C_{r} a^{n-r }b^{r}

Assuming that x^{5 }occurs in the (r + 1)th term of the expansion (x + 3)^{8}, we obtain

T_{r+1 }^{ }= ^{8}C_{r} (x)^{8-r}(3)^{r}

Comparing the indices of x in x^{5} and in T_{r +1},

we obtain r = 3

Thus, the coefficient of x^{5} is ^{8}C_{3}(3)^{3} = ^{8!}/_{3!5!} x 3^{3} = ^{8.7.6.5!}/_{3.2.5!} .3^{3} = 1512

**Find the coefficient of a**^{5}b^{7}in (a – 2b)^{12}

Solution:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n }is given by T_{r+1} = ^{n}C_{r} a^{n-r }b^{r}

Assuming that a^{5}b^{7} occurs in the (r + 1)^{th} term of the expansion (a – 2b)^{12}, obtain

T_{r+1} = ^{12}C_{r}(a)^{12-r}(-2b)^{r} = ^{12}C_{r} (-2)^{r}(a)^{12-r}(b)^{r}

Comparing the indices of a and b in a^{5}b^{7} and in T_{r+1}

we obtain r = 7

Thus the coefficient of a^{5}b^{7} is

^{12}C_{7}(-2)^{7} = –^{12!}/_{7!5!} .2^{7} = –^{12.11.10.9.8.7!}/_{5.4.3.2.7!} .2^{7} = -(792)(128) = -101376

**Write the general term in the expansion of (x**^{2}– y)^{6}

Solution:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n }is given by T_{r+1} = ^{n}C_{r} a^{n-r }b^{r}

Thus the general term in the expansion of (x^{2} – y)^{6} is

T_{r+1} = ^{6}C_{r}(x^{2})^{6-r}(-y)^{r} = (-1)^{r} ^{6}C_{r} (x)^{12-2r}(y)^{r}

**Write the general term in the expansion of (x**^{2}– yx)^{12}, x≠0

Solution:

^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n }is given by T_{r+1} = ^{n}C_{r} a^{n-r }b^{r}

Thus the general term in the expansion of (x^{2} – yx)^{12}is

T_{r+1} = ^{12}C_{r}(x^{2})^{12-r}(-yx)^{r} = (-1)^{r} ^{12}C_{r} (x)^{24-r}(y)^{r}

**Find the 4**^{th}term in the expansion of (x – 2y)^{12}

Solution:

^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n }is given by T_{r+1} = ^{n}C_{r} a^{n-r }b^{r}

Thus, the 4^{th} term in the expansion of (x – 2y)^{12}

T_{4} = T_{3+1} = ^{12}C_{3} (x)^{12-3}(-2y)^{3} = (-1)^{3} ^{12!}/_{3!9!} .x^{9}.(2)^{3}.y^{3} = – ^{12.11.10}/_{3.2} .(2)^{3}.x^{9}y^{3} = -1760x^{9}y^{3}

**Find the 13**^{th}term in the expansion of (9x –^{1}/_{3√x})^{18}, x≠0

Solution:

It is known that (r+1)^{th} term, (T_{r+1}) in the binomial expansion of (a+b)n is given by T_{r+1} = ^{n}C_{r} a^{n-r }b^{r}

The 13^{th} term in the expansion of (9x – ^{1}/_{3√x})^{18}, x≠0

T_{13} = T_{12+1} = ^{18}C_{12} (9x)^{18-12 }(– ^{1}/_{3√x})^{12}

=(-1)^{12} . ^{18!}/_{12!6!} (9)^{6}(x)^{6} (^{1}/_{3})^{12}(^{1}/_{√x})^{12}

= ^{18.17.16.15.14.13.12!}/_{12!6.5.4.3.2.} x^{6}.(^{1}/_{x}^{6}).3^{12}.(^{1}/_{3}^{12})

= 18564

**Find the middle terms in the expansion of [3 – (x**^{3}/_{6})]^{7}

Solution:

It is known that in the expansion of (a+b)^{n}, if n is odd, then there are two middle terms, namely (^{n+1}/_{2})^{th} term and (^{n+1}/_{2} + 1)^{th} term.

Therefore the middle terms in the expansion [3 – (x^{3}/_{6})]^{7} are (^{7+1}/_{2})^{th} = 4^{th} term and (^{7+1}/_{2} + 1)^{th} = 5^{th} term

T_{4} = T_{3+1} = ^{7}C_{3} (3)^{7-3} [-(x^{3}/_{6})]^{3} = (-1)^{3} ^{7!}/_{3!4!} . 3^{4}.(x^{9}/_{6}^{3})

= –^{7.6.5.4!}/_{3.2.4!} . 3^{4}.[^{1}/_{2}^{3}._{3}^{3}].x^{9} = –^{105}/_{8} .x^{9}

T_{5} = T_{4+1} = ^{7}C_{4} (3)^{7-4} [-(x^{3}/_{6})]^{4} = (-1)^{4} ^{7!}/_{3!4!} . 3^{3}.(x^{12}/_{6}^{4})

= –^{7.6.5.4!}/_{3.2.4!} . [(3^{3})/_{2}^{4}._{3}^{4}].x^{12} = ^{35}/_{48} .x^{12}

Thus, the middle term in the expansion of [3 – (x^{3}/_{6})]^{7} are –^{105}/_{8}.x^{9} and ^{35}/_{48}.x^{12}

**Find the middle term in the expansion of (**^{x}/_{3}+ 9y)^{10}

Solution:

It is known that in the expansion (a+b)^{n}, if n is even, then the middle term is (^{n}/_{2} + 1)^{th} term.

Therefore the middle term in the expansion of (^{x}/_{3} + 9y)^{10} is (^{10}/_{2} + 1)^{th} term

Therefore, the middle term in the expansion of (^{x}/_{3} + 9y)^{10} is (^{10}/_{2} + 1)^{th} = 6^{th}

T_{6} = T_{5+1} = ^{10}C_{5} (^{x}/_{3})^{10-5} (9y)^{5} =^{10!}/_{5!5!} . 9^{5}.(x^{5}/_{3}^{5}).y^{5}

= –^{10.9.8.7.6.5!}/_{5.4.3.2.5!} . [^{1}/_{3}^{5}].3^{10}.x^{5}.y^{5 }

=252.3^{5}.x^{5}.y^{5}

Thus, the middle term in the expansion of (^{x}/_{3} + 9y)^{10} is 61236x^{5}y^{5}

**In the expansion of (1 + a)**^{m+n}prove that coefficient of a^{m}and a^{n}are equal

Solution:

It is known that (r+1)^{th} term, (T_{r+1}) in the binomial expansion of (a+b)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Assuming that a^{m} occurs in the (r+1)_{th }term, (T_{r+1}) in the binomial expansion of (a+b)^{n} is given by T_{r+1} = ^{n}C_{r }a^{n-r} b^{r}

Assuming that a^{m} occurs in the (r+1)^{th} term of the expansion (1+a)^{m+n}, we obtain

T_{r+1} = ^{m+n}C_{r}(1)^{m+n-r}(a)^{r} = ^{m+n}C_{r}.a^{r}

Comparing the indices of a in a^{m} and in T_{r+1}, we obtain

r = m

Therefore the coefficient of a^{m} is

^{m+n}C_{m} = ^{(m+n)!}/_{m!(m+n-m)!} = ^{(m+n)!}/_{m!n!} ———–(1)

Assuming that a^{n} occurs in the (k+1)^{th} term of the expansion (1 + a)^{m+n} we obtain

T_{k+1} = ^{m+n}C_{k}(1)^{m+n-k}(a)^{k} = ^{m+n}C_{k}(a)^{k}

comparing th indices of a and a^{n} and in T_{k+1}

we obtain

k = n

herefore the coefficient of a^{n} is

^{m+n}C_{n} = ^{(m+n)!}/_{n!(m+n-m)!} =^{(m+n)!}/_{n!m! }——————-(2)

Thus from (1) and(2) it can be observed that the coefficients of a^{m} and a^{n} in the expansion of (1+a)^{m+n} are equal.

**The coefficients of the (r-1)**^{th}, r^{th}and (r+1)^{th}terms in the expansion of (x+1)^{n}are in the ratio 1:3:5. Find n and r.

Solution:

It is known that (k+1)^{th} term, (T_{k+1}) in the binomial expansion of (a+b)^{n} is given by T_{k+1} = ^{n}C_{r} a^{n-k }b^{k}

Therefore (r – 1)^{th} term in the expansion of (x+1)^{n} is

T_{r-1} = ^{n}C_{r-2} (x)^{n-(r-2)}(1)^{r-2} = ^{n}C_{r-2} x^{n-r+2}

(r+1) term in the expansion of (x+1)^{n} is

T_{r+1} = ^{n}C_{r}(x)^{n-r}(1)^{r} = ^{n}C_{r} x^{n-r}

r^{th} term in the expansion of (x + 1)^{n} is

T_{r} = ^{n}C_{r -1 }(x)^{n-(r-1)}(1)^{(r-1)} = ^{n}C_{r-1 }x^{n-r+1}

Therefore the coefficients of the (r -1)^{th}, r^{th} and (r +1)^{th} terms in the expansion of (x + 1)^{n} ^{n}C_{r -2}, ^{n}C_{r-1}^{ and n}C_{r} are respectively. Since these coefficients are in the ratio 1:3:5, we obtain

^{n}C_{r-2}/^{n}C_{r-1} = ^{1}/_{3} and ^{n}C_{r-1}/^{n}C_{r} = ^{3}/_{5}

^{n}C_{r-2}/^{n}C_{r-1} = ^{n!}/_{(r-2)!(n-r+2)!} x ^{(r-1)(n-r+1)!}/_{n!} = ^{(r-1)(r-2)(n-r+1)!}/^{ !}/_{(r-2)!(n-r+2)!(n-r+1)!}

= ^{r-1}/_{n-r+2}

^{r-1}/_{n-r+2} = ^{1}/_{3}

3r – 3 = n – r + 2

n – 4r + 5 = 0

^{n}C_{r-1}/^{n}C_{r} =^{ n!}/_{(r-1)!(n-r+1)!} x ^{(r!)(n-r)!}/_{n!} = ^{r(r-1)!(n-r)!}/_{(r-1)!(n-r+1)(n-r)!}

=^{r}/_{n-r+1}

^{r}/_{n-r+1} = ^{3}/_{5}

5r = 3n – 3r + 3

3n – 8r + 3 = 0

Multiplying (1) by 3 and subtracting it from (2), we obtain

4r – 12 = 0

⇒ r = 3

Putting the value of r in (1), we obtain n

– 12 + 5 = 0

⇒ n = 7

Thus, n = 7 and r = 3

**Prove that the coefficient of x**^{n}in the expansion of (1 + x)^{2n}is twice the coefficient of x^{n}in the expansion of (1 + x)^{2n–1}

Solution:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{2n}C_{r} a^{n-r} b^{r}

Assuming that x^{n} occurs in the (r + 1)^{th} term of the expansion of (1 + x)^{2n}, we obtain

T_{r+1} = ^{2n}C_{r} (1)^{2n-r} (x)^{r} = ^{2n}C_{r} (x)^{r}

Comparing the indices of x in x^{n} and in T^{r + 1}, we obtain r = n

Therefore, the coefficient of x^{n} in the expansion of (1 + x)^{2n} is

^{2n}C_{n} = ^{(2n)!}/_{n!(2n-n)! }= ^{(2n)!}/_{n!n!} = ^{(2n)!}/_{(n!)}^{2}———–(1)

Assuming that x^{n} occurs in the (k +1)^{th }term of the expansion (1 + x)^{2n – 1}, we obtain

T_{r+1} = ^{2n-1}C_{k} (1)^{2n-1-k} (x)^{k} = ^{2n-1}C_{k} (x)^{k}

Comparing the indices of x in x^{n} and T_{k + 1}, we obtain k = n

Therefore, the coefficient of x^{n} in the expansion of (1 + x)^{2n –1} is

^{2n-1}C_{n} = ^{(2n-1)!}/_{n!(2n-1-n)! }= ^{(2n-1)!}/_{n!(n-1)!} _{)}

= ^{2n.(2n-1)!}/_{2n.n!(n-1)!} = ^{(2n)!}/_{2.n!n!} = ^{1}/_{2}[^{(2n)!}/_{(n!)}^{2}] ————(2)

From (1) and (2) it is observed that

^{1}/_{2} (^{2n}C_{n}) =^{2n-1}C_{n}

^{2n}C_{n} =2(^{2n-1}C_{n})

Therefore, the coefficient of x^{n} in the expansion of (1 + x)^{2n} is twice the coefficient of x^{n} in the expansion of (1 + x)^{2n–1}.

Hence, proved.

**Find a positive value of m for which the coefficient of x**^{2 }in the expansion (1 + x)^{m}is 6.

Solution:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{2n}C_{r} a^{n-r} b^{r}

Assuming that x^{2} occurs in the (r + 1)^{th} term of the expansion of (1 + x)^{m}, we obtain

T_{r+1} = ^{m}C_{r} (1)^{m-r} (x)^{r} = ^{m}C_{r} (x)^{r}

Comparing the indices of x in x^{2} and in T^{r + 1}, we obtain r = 2

Therefore, the coefficient of x^{2} is ^{m}C_{2} in the expansion of (1 + x)^{m} is 6.

^{m}C_{2} = 6

^{(m)!}/_{2!(m-2)! }=6

^{(m)(m-1)(m-2)!}/_{2x(m-2)!} = 6

m(m – 1) = 12

m^{2} – m – 12 = 0

m^{2} – 4m + 3m – 12 = 0

m(m-4)+3(m – 4) = 0

(m – 4)(m + 3) = 0

(m – 4) = 0 or (m + 3) = 0

m = 4 or m = -3

Thus, the positive value of m for which the coefficient of x^{2} in the expansion (1+ x)^{m} is 6, is 4.