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If the coefficients of rth and (r + 1)th terms in the expansion of (3 + 7x)29 are equal, then r equals

Option: 1

15 


Option: 2

21 


Option: 3

14 


Option: 4

none of these 


Answers (1)

best_answer

As we have learned

Comparing of Coefficients -

T_{r+1}=^{n} c_{r}\cdot x^{n-r}\left ( f\left ( r \right ) \right )^{2}

We arrange all of x together and make x^{\left ( n,\,r \right )}

compare :x^{\left ( n\,c,r \right )}= x^{R}

find r

T_{r+1}=^{n} c_{r}\cdot x^{n-r}\left ( f\left ( r \right ) \right )^{2}

We arrange all of x together and make x^{\left ( n,\,r \right )}

compare :x^{\left ( n\,c,r \right )}= x^{R}

find r

 

- wherein

r can not be -ve or fractions .

 

 

We have,                                                                                                 

                             T_{r+1} = ^{29}C_r 3^{29-r } (7x )^r = (^{29}C_r \times 3 ^{29-r}\times 7^r )x^r

                         ar = coefficient of (r + 1)th term = 29Cr´ 329-r´ 7r

                    and    ar-1 = coefficient of rth term = 29Cr-1´ 330-r´ 7r-1

                    Now,  ar = ar-1

                         ^{29}C_r \times 3^{29-r } \times (7 )^r = (^{29}C_{r-1} \times 3 ^{30-r}\times 7^{r-1} )

                          \frac{^{29}C_r }{^{29}C_{r-1}}= 3/7 \Rightarrow \frac{30 -r }{r } = 3/7 \Rightarrow r = 21 

 

Posted by

Riya

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