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Miscellaneous

Binomial Theorem

**Example-1 :-** Find the term independent of x in the expansion of (3x^{2}/2 - x/3)^{6}.

We have, T_{r+1}=^{6}C_{r}(3x^{2}/2)^{6-r}(-1/3x)^{r}=^{6}C_{r}(3/2)^{6-r}(x^{2})^{6-r}(-1)^{r}(1/x)^{r}(1/3)^{r}= (-1)^{r}^{6}C_{r}[3^{6-2r}/2^{6-r}] x^{12-3r}The term will be independent of x if the index of x is zero, i.e., 12 – 3r = 0. Thus, r = 4 Hence 5th term is independent of x and is given by (-1)^{4}^{6}C_{4}[3^{6-8}/2^{6-4}] x^{12-12}= 5/12

**Example-2 :-** If the coefficients of a^{r-1}, a^{r} and a^{r+1} in the expansion of (1 + a)^{n} are in arithmetic progression, prove that n^{2} – n(4r + 1) + 4r^{2} – 2 = 0.

The (r + 1)th term in the expansion is nCrar. Thus it can be seen that ar occurs in the (r + 1)th term, and its coefficient is^{n}C_{r}. Hence the coefficients of a^{r-1}, a^{r}and a^{r+1}are^{n}C_{r-1},^{n}C_{r}and^{n}C_{r+1}, respectively. Since these coefficients are in arithmetic progression, so we have,^{n}C_{r-1}+^{n}C_{r+1}= 2.^{n}C_{r}. This gives r(r + 1) + (n – r) (n – r + 1) = 2 (r + 1) (n – r + 1) or r^{2}+ r + n^{2}– nr + n – nr + r^{2}– r = 2(nr – r^{2}+ r + n – r + 1) n^{2}– 4nr – n + 4r^{2}– 2 = 0 n^{2}– n (4r + 1) + 4r^{2}– 2 = 0

**Example-3 :-** Show that the coefficient of the middle term in the expansion of (1 + x)^{2n} is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)^{2n-1}.

As 2n is even so the expansion (1 + x)^{2n}has only one middle term which is (2n/2 + 1)th i.e., (n + 1)th term. The (n + 1)th term is^{2n}C_{n}x^{n}. The coefficient of x^{n}is^{2n}C_{n}Similarly, (2n – 1) being odd, the other expansion has two middle terms, [(2n-1+1)/2]th and [(2n-1+1)/2 + 1]th i.e., nth and (n + 1)th terms. The coefficients of these terms are^{2n-1}C_{n-1}and^{2n-1}C_{n}, respectively. Now,^{2n-1}C_{n-1}+^{2n-1}C_{n}=^{2n}C_{n}[As^{n}C_{r-1}+^{n}C_{r}=^{n+1}C_{r}]. as required.

**Example-4 :-** Find the coefficient of a^{4 in the product (1 + 2a)4 (2 – a)5 using binomial theorem. }

We first expand each of the factors of the given product using Binomial Theorem. We have (1 + 2a)^{4}=^{4}C_{0}+^{4}C_{1}(2a) +^{4}C_{2}(2a)^{2}+^{4}C_{3}(2a)^{3}+^{4}C_{4}(2a)^{4}= 1 + 4 (2a) + 6(4a^{2}) + 4 (8a^{3}) + 16a^{4}. = 1 + 8a + 24a^{2}+ 32a^{3}+ 16a^{4}Now, (2 – a)^{5}=^{5}C_{0}(2)^{5}–^{5}C_{1}(2)^{4}(a) +^{5}C_{2}(2)^{3}(a)^{2}–^{5}C_{3}(2)^{2}(a)^{3}+^{5}C_{4}(2) (a)^{4}–^{5}C_{5}(a)^{5}= 32 – 80a + 80a^{2}– 40a^{3}+ 10a^{4}– a^{5}Thus (1 + 2a)^{4}(2 – a)^{5}= (1 + 8a + 24a^{2}+ 32a^{3}+ 16a^{4}) (32 – 80a + 80a^{2}– 40a^{3}+ 10a^{4}– a^{5}) The complete multiplication of the two brackets need not be carried out. We write only those terms which involve a^{4}. This can be done if we note that a^{r}. a^{4-r}= a^{4}. The terms containing a^{4}are 1 (10a^{4}) + (8a) (–40a^{3}) + (24a^{2}) (80a^{2}) + (32a^{3}) (– 80a) + (16a^{4}) (32) = – 438a^{4}Thus, the coefficient of a^{4}in the given product is – 438.

**Example-5 :-** Find the rth term from the end in the expansion of (x + a)^{n}.

There are (n + 1) terms in the expansion of (x + a)^{n}. Observing the terms we can say that the first term from the end is the last term, i.e., (n + 1)th term of the expansion and n + 1 = (n + 1) – (1 – 1). The second term from the end is the nth term of the expansion, and n = (n + 1) – (2 – 1). The third term from the end is the (n – 1)th term of the expansion and n – 1 = (n + 1) – (3 – 1) and so on. Thus rth term from the end will be term number (n + 1) – (r – 1) = (n – r + 2) of the expansion. And the (n – r + 2)th term is^{n}C_{n–r+1}x^{r-1}a^{n-r+1}.

**Example-6 :-** Find the term independent of x in the expansion of

Since we have to find a term independent of x, i.e., term not having x, so take (18 - 2r)/3 = 0 We get r = 9. The required term is^{18}C_{9}1/2^{9}

**Example-7 :-** The sum of the coefficients of the first three terms in the expansion of (x - 3/x^{2})^{m} , x ≠ 0, m being a natural number, is 559. Find the term of the expansion containing x_{3}.

The coefficients of the first three terms of (x - 3/x^{2})^{m}are^{m}C_{0}, (–3)^{m}C_{1}and 9^{m}C_{2}. Therefore, by the given condition, we have^{m}C_{0}(–3)^{m}C_{1}+ 9^{m}C_{2}= 559, i.e., 1 – 3m + [9m(m-1)/2] = 559 which gives m = 12 (m being a natural number). Now, T_{r+1}=^{12}C_{r}x^{12-r}(-3/x^{2})^{r}=^{12}C_{r}(–3)^{r}. x^{12-3r}Since we need the term containing x^{3}, so put 12 – 3r = 3 i.e., r = 3. Thus, the required term is^{12}C_{3}(–3)^{3}x^{3}, i.e., – 5940 x^{3}.

**Example-8 :-** If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)^{34} are equal, find r.

The coefficients of (r – 5)th and (2r – 1)th terms of the expansion (1 + x)^{34}are^{34}C_{r-6}and^{34}C_{2r-2}, respectively. Since they are equal so,^{34}C_{r-6}=^{34}C_{2r-2}Therefore, either r – 6 = 2r – 2 or r – 6 = 34 – (2r – 2) [Using the fact that if^{n}C_{r}=^{n}C_{p}, then either r = p or r = n – p] So, we get r = – 4 or r = 14. r being a natural number, r = – 4 is not possible. So, r = 14.

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