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Find the coefficient of

    Q1.    x^5 in (x + 3)^8

Answers (1)

As we know that the (r+1)^{th} term  T_{r+1} in the binomial expansion of  (a+b)^n  is given by 

T_{r+1}=^nC_ra^{n-r}b^r

Now let's assume x^5 happens in the (r+1)^{th} term of the binomial expansion of (x + 3)^8

So,

T_{r+1}=^8C_rx^{8-r}3^r

On comparing the indices of x we get,

r=3

Hence the coefficient of the  x^5 in (x + 3)^8 is 

^8C_3\times3^3=\frac{8!}{5!3!}\times 9=\frac{8\times7\times6}{3\times2}\times9=1512

Posted by

neha

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