solve this

if the coefficients of (r-5)^{th} and (2r-1)^{th} term in the expansion of (1+x)^{34} are equal, fiind r

We know that the coefficient of *r*^{th} in the expansion of (1 + *x*)* ^{n}* is

*C*

^{n}

_{r}_{ – 1}.

∴ Coefficients of (*r* – 5)^{th} and (2*r* – 1)^{th} term in the Expansion of (1 + *x*)^{34} are ^{34}C_{r}_{ – 6} and ^{34}C_{2}_{r}_{ – 2}

It is given that these coefficients are equal

^{34}C_{r}_{ – 6} = ^{34}C_{2}_{r}_{ – 2}

⇒ *r* – 6 = 2*r* – 2 or *r* – 6 + 2*r* – 2 = 34 (* ^{n}*C

*=*

_{r}*C*

^{n}_{S}⇒

*r*= S or

*r*+ S =

*n*)

⇒ 3*r* – 8 = 34 (*r* – 6 = 2*r* – 2 ⇒ *r* = –4 which is not possible)

⇒ 3*r* = 34 + 8 = 42

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