Binomial Theorem
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4 months agoContributor-Level 10
5. Let (r + 1)th term be the general term of (x2–y)6.
So, Tr-1 = 6Cr (x2)6-r (-y)r
= (–1)r .6Cr . .
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4 months agoContributor-Level 10
4. Let a5b7 occurs in (r + 1)th term of [removed]a – 2b)12.
Now, Tr+1 = 12Cra12-r (–2b)r
= (–1)r12Cra12–r . 2r. br
Comparing indices of a and b in Tr-1 with a5 and b7 we get, r = 7
So, co-efficient of a5b7 is (–1)712C7 27
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4 months agoNew question posted
4 months agoNew answer posted
4 months agoContributor-Level 10
2. 
By binomial theorem,
(a + b)n = nC0 (a)n (b)0 + nC1 (a)n–1 (b)1 + …………. + nCr (a)n–r (b)r + …………… + nCn (a)n–n (b)n
Where, b0 = 1 = an–n
So, (a + b)n = nCr (a)n–r (b)r
Putting a = 1 and b = 3 such that a + b = 4, we can rewrite the above equation as
(1 + 3)n = nCr (1)n–r.3r
=>4n = .nCr
Hence proved.
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4 months agoContributor-Level 10
1. For a number x to be divisible by y, we can write x as a factor of y i.e., x = ky where k is some natural number. Thus in order for 9n+1 – 8n – 9 to be divisible by 64 we need to show that 9n+1 – 8n – 9 = 64k where k is some natural number.
We have, by binomial theorem
(1 + a)m = mC0 + mC1 (a) + mC2 (a)2 + mC3 (a)3 + ………… + mCm (a)m
Putting, a = 8 and m = n + 1
(1 + 8)n+1 = n+1C0 + n+1C1.8 + n+1C2.82 + n+1C3.83 + ……. + n+1Cn+1. (8)n+1
=> 9n+1= 1 + (n + 1)8 + 82* [n+1C2 + n+1C3.8 + ………. + n+1Cn+1. (8)n+1–2] [since, n+1C0 = 1, n+1C1= n + 1]
=> 9n+1 = 1 + 8n + 8 + 64 * [n+1C2 + n+1C3.8 + ……
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