Write down the general term of the binomial expansion of (2x+3/x)^14

Show that there are exactly two consecutive terms with equal coefficients.

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Given the binomial `(2x + 3/x)^14`

This is a binomial of the type `(x+y)^n`

The general term, i.e. the *(r+1)th* term is given by:

`t_(r+1) = ^(14)c_r*(2x)^(14-r)*(3/x)^r `

`= ^(14)c_r*2^(14-r)*3^r*x^(14-r-r)`

`= ^(14)c_r*2^(14-r)*3^r*x^(14-2r)`

Now, ratio of two consecutive terms,

` ` `(t_(r+1))/(t_r) = (^14c_r*(2)^(14-r)*(3)^r*x^(14-2r))/(^14c_(r-1)*2^(14-(r-1))*3^(r-1)*x^(14-2(r-1)))`

`=(^14c_r)/(^14c_(r-1))*((3)^1)/((2)^1*x^2)`

`= (14-r+1)/(r)*(3/2)*1/x^2`

`= (15-r)/r*(3/2)*1/x^2`

Therefore, the ratio of the co-efficients of the (r+1)th term to rth term will be `=(15-r)/r*(3/2)`

When the co-efficients of two consecutive terms are equal, the ratio `c_1/c_2` will be = 1.

So, `(15-r)/r*3/2 = 1`

or, 45 - 3r = 2r

or, 5r = 45,

or, r = 9

**As this is the only solution of r, there will be exactly two consecutive terms whose co-efficients are same. They are the 9th and 10th terms in the series.**

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