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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 `
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)))`
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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