Q15 of 25 Page 8

Prove that x5 in (x + 3)8


Suppose x5 occurs in the (r + 1 )th term of the expansion (x + 3)8
Now Tr+1 = nCr an - r br = 8Cr x8 - r 3r
Comparing the indices of x in x5 and in Tr + 1, we get r = 3
Thus, the coefficient of x5 is
8C3(3)3 = 1512

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