TOPICS

Unit-8(Examples)

Binomial Theorem

**Example-1 :-** Expand (x^{2} + 3/x)^{4}, x ≠ 0

(x^{2}+ 3/x)^{4}=^{4}C_{0}(x^{2})^{4}+^{4}C_{1}(x^{2})^{3}(3/x) +^{4}C_{2}(x^{2})^{2}(3/x)^{2}+^{4}C_{3}(x^{2})(3/x)^{3}+^{4}C_{4}(3/x)^{4}= x^{8}+ 4x^{6}. 3/x + 6x^{4}. 9/x^{2}+ 4x^{2}. 27/x^{3}+ 81/x^{4}= x^{8}+ 12x^{5}+ 54x^{2}+ 108/x + 81/x^{4}

**Example-2 :-** Compute (98)^{5}.

98^{5}= (100 - 2)^{5}=^{5}C_{0}(100)^{5}-^{5}C_{1}(100)^{4}.2 +^{5}C_{2}(100)^{3}.2^{2}-^{5}C_{3}(100)^{2}.2^{3}+^{5}C_{4}(100).2^{4}-^{5}C_{5}(2)^{5}= 100000000000 - 5 x 100000000 x 2 + 10 x 1000000 x 4 - 10 x 10000 x 8 + 5 x 100 x 16 - 32 = 10040008000 - 1000800032 = 9039207968

**Example-3 :-** Which is larger (1.01)^{1000000} or 10,000?

(1.01)^{1000000}= (1 + 0.01)^{1000000}=^{1000000}C_{0}+^{1000000}C_{1}(0.01) + other positive terms = 1 + 1000000 × 0.01 + other positive terms = 1 + 10000 + other positive terms > 10000 Hence (1.01)^{1000000}> 10000

**Example-4 :-** Using binomial theorem, prove that 6^{n} – 5^{n} always leaves remainder 1 when divided by 25.

For two numbers a and b if we can find numbers q and r such that a = bq + r, then we say that b divides a with q as quotient and r as remainder. Thus, in order to show that 6^{n}– 5^{n}leaves remainder 1 when divided by 25, we prove that 6^{n}– 5^{n}= 25k + 1, where k is some natural number. We have (1 + a)^{n}=^{n}C_{0}+^{n}C_{1}a +^{n}C_{2}a^{2}+ ... +^{n}C_{n}a^{n}For a = 5, we get (1 + 5)^{n}=^{n}C_{0}+^{n}C_{1}5 +^{n}C_{2}5^{2}+ ... +^{n}C_{n}5^{n}i.e. (6)^{n}= 1 + 5n + 5^{2}.^{n}C_{2}+ 5^{3}.^{n}C_{3}+ ... + 5^{n}i.e. 6^{n}– 5^{n}= 1 + 5^{2}(^{n}C_{2}+^{n}C_{3}5 + ... + 5^{n-2}) or 6^{n}– 5^{n}= 1 + 25(^{n}C_{2}+ 5 .^{n}C_{3}+ ... + 5^{n-2}) or 6^{n}– 5^{n}= 25k + 1 where k =^{n}C_{2}+ 5 .^{n}C_{3}+ ... + 5^{n-2}. This shows that when divided by 25, 6^{n}– 5^{n}leaves remainder 1.

**Example-5 :-** Find a if the 17th and 18th terms of the expansion (2 + a)^{50} are equal.

The (r + 1)th term of the expansion (x + y)^{n}is given by T_{r+1}=^{n}C_{r}x^{n-r}y^{r}. For the 17th term, we have, r + 1 = 17, i.e., r = 16 Therefore, T_{17}= T_{16+1}=^{50}C_{16}(2)^{50–16}a^{16}=^{50}C_{16}2^{34}a^{16}. Similarly, T_{18}=^{50}C_{17}2^{33}a^{17}Given that T_{17}= T_{18}So,^{50}C_{16}(2)^{34}a^{16}=^{50}C_{17}(2)^{33}a^{17}Therefore, [^{50}C_{16}(2)^{34}] ÷ [^{50}C_{17}(2)^{33}] = a^{17}÷ a^{16}a = 50!/(16!34!) x (17!33!)/50! x 2 = 1

**Example-6 :-** Show that the middle term in the expansion of (1+x)^{2n} is [1.3.5...(2n-1)]/n! 2n x^{n}, where n is a positive integer.

As 2n is even, the middle term of the expansion (1 + x)^{2n}is (2n/2 + 1)th, i.e., (n + 1)th term which is given by,

**Example-7 :-** Find the coefficient of x^{6}y^{3} in the expansion of (x + 2y)^{9}.

Suppose x^{6}y^{3}occurs in the (r + 1)th term of the expansion (x + 2y)^{9}. Now T_{r+1}=^{9}C_{r}x^{9-r}(2y)^{r}=^{9}C_{r}2^{r}. x^{9-r}. y^{r}. Comparing the indices of x as well as y in x^{6}y^{3}and in T_{r+1}, we get r = 3. Thus, the coefficient of x^{6}y^{3}is^{9}C_{3}2^{3}= 9!/(3!6!) x 2^{3}= 672

**Example-8 :-** The second, third and fourth terms in the binomial expansion (x + a)^{n} are 240, 720 and 1080, respectively. Find x, a and n.

Given that second term T_{2}= 240 We have T_{2}=^{n}C_{1}x^{n-1}. a So^{n}C_{1}x^{n-1}. a = 240 ... (1) Similarly^{n}C_{2}x^{n-2}a^{2}= 720 ... (2) and^{n}C_{3}x^{n-3}a^{3}= 1080 ... (3) Dividing (2) by (1), we get [^{n}C_{2}x^{n-2}a^{2}] ÷ [^{n}C_{1}x^{n-1}. a] = 720 ÷ 240 [(n-1)!.a] ÷ [(n-2)!.x] = 6 ....(4) a/x = 6/(n-1) Dividing (3) by (2), we have [^{n}C_{3}x^{n-3}a^{3}] ÷ [^{n}C_{2}x^{n-2}a^{2}] = 1080 ÷ 720 a/x = 9/[2(n-2)] ....(5) From (4) and (5), 6/(n-1) = 9/[2(n-2)] 9n - 9 = 12n - 24 12n - 9n = 24 - 9 3n = 15 n = 5 Hence, from (1), 5x^{4}a = 240, and from (4), a/x = 3/2 Solving these equations for a and x, we get x = 2 and a = 3.

**Example-9 :-** The coefficients of three consecutive terms in the expansion of (1 + a)^{n} are in the ratio1: 7 : 42. Find n.

Suppose the three consecutive terms in the expansion of (1 + a)^{n}are (r – 1)th, rth and (r + 1)th terms. The (r – 1)th term is^{n}C_{r-2}a^{r-2}, and its coefficient is^{n}C_{r-2}. Similarly, the coefficients of rth and (r + 1)th terms are^{n}C_{r-1}and^{n}C_{r}, respectively. Since the coefficients are in the ratio 1 : 7 : 42, so we have,^{n}C_{r-2}:^{n}C_{r-1}= 1/7, i.e., n – 8r + 9 = 0 ... (1)^{n}C_{r-1}:^{n}C_{r}= 7/42 , i.e., n – 7r + 1 = 0 ... (2) Solving equations(1) and (2), we get, n = 55.

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