Let us study the concept by finding the constant term in the expansion of (x3 – 1/x2)15

In (a + b)n the (r + 1)th term Tr+1 is given by –
Tr+1 = nCr .an-r. br for r = 0, 1, 2, 3, 4, 5 … n

Let constant terms occurs at (r+1)th term for ( x3 – 1/x2 )15
So,
Tr+1 = nCr an-r.br =  15Cr (x3)15-r.(x-2)r
Tr+1 = 15Cr (x)45-5r

For constant term, x  should be eliminated, so –
45 -5r = 0
r = 9

So the constant term is T9+1 , so T10

T10  = 15C9 (x)45-45
T10  = 15C9

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