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Solved Examples on Binomial Theorem Illustration 1: In the binomial expansion of (a-b)n, n ≥ 5the sum of the 5th and 6th terms is zero. Then what is the value of a/b? (2001) Solution: Let us denote the fifth term as T5 and the sixth term as T6. So, it is given that T5 + T6 = 0 This gives nC4an-4 b4 - nC5an-5 b5 = 0 Hence, nC4an-4 b4 = nC5an-5 b5 In order to obtain the value of a/b, we shift the terms obtained above on one side. This gives the value of a/b as = nC5/ nC4 = (n-4)/5. Illustration 2: Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1- Tn = 21, then what is the value of n? (2001) Solution: According to the given condition, Tn = nC3 Tn+1- Tn = 21 Hence,n+1C3- nC3 = 21 So, (n+1)(n-1)n/6 – n(n-1)(n-2)/6 = 21 So, n(n-1)3/6 = 21 Hence, n(n-1) = 42 This gives, n =7. Illustration 3: Coefficient of t24 in (1 + t2)12(1 + t12)(1 + t24) is….? (2003) Solution: Here, coefficient of t24 in (1 + t2)12(1 + t12)(1 + t24) This is same as the coefficient of t24 in (1 + t2)12(1 + t12 + t24 + t36) Or the coefficient of t24 in (1 + t2)12+t12 (1 + t2)12+ t24 (1 + t2)12 Hence,the coefficient of t24 in 12C12 + 12C6 + 12C0 = 2 + 12C6
Illustration 1: In the binomial expansion of (a-b)n, n ≥ 5the sum of the 5th and 6th terms is zero. Then what is the value of a/b? (2001)
Solution: Let us denote the fifth term as T5 and the sixth term as T6.
So, it is given that T5 + T6 = 0
This gives nC4an-4 b4 - nC5an-5 b5 = 0
Hence, nC4an-4 b4 = nC5an-5 b5
In order to obtain the value of a/b, we shift the terms obtained above on one side.
This gives the value of a/b as = nC5/ nC4 = (n-4)/5.
Illustration 2:
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1- Tn = 21, then what is the value of n? (2001)
Solution: According to the given condition, Tn = nC3
Tn+1- Tn = 21
Hence,n+1C3- nC3 = 21
So, (n+1)(n-1)n/6 – n(n-1)(n-2)/6 = 21
So, n(n-1)3/6 = 21
Hence, n(n-1) = 42
This gives, n =7.
Illustration 3: Coefficient of t24 in (1 + t2)12(1 + t12)(1 + t24) is….? (2003)
Solution: Here, coefficient of t24 in (1 + t2)12(1 + t12)(1 + t24)
This is same as the coefficient of t24 in (1 + t2)12(1 + t12 + t24 + t36)
Or the coefficient of t24 in (1 + t2)12+t12 (1 + t2)12+ t24 (1 + t2)12
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