Find the value of r, If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal.

As an experienced tutor registered on UrbanPro, I can guide you through this problem efficiently. Let's tackle it step by step. Firstly, we need to understand that when expanding (1+x)34(1+x)34 using the binomial theorem, the rrth term is represented by (34r)xr(r34)xr. So, the coefficient of the (r−5)(r−5)th term would be (34r−5)(r−534), and the coefficient of the (2r−1)(2r−1)th term would be (342r−1)(2r−134). Given that these coefficients are equal, we can set up the equation: (34r−5)=(342r−1)(r−534)=(2r−134) To solve this equation, we can use the formula for binomial coefficients: (nk)=n!k!(n−k)!(kn)=k!(n−k)!n! Substituting this into our equation, we get: 34!(r−5)!(34−(r−5))!=34!(2r−1)!(34−(2r−1))!(r−5)!(34−(r−5))!34!=(2r−1)!(34−(2r−1))!34! Now, we can cancel out the 34!34! from both sides of the equation: (34−(r−5))!(r−5)!=(34−(2r−1))!(2r−1)!(r−5)!(34−(r−5))!=(2r−1)!(34−(2r−1))! (39−r)!(r−5)!=(35−2r)!(2r−1)!(r−5)!(39−r)!=(2r−1)!(35−2r)! Now, we need to simplify further: (39−r)(38−r)(37−r)(36−r)(35−r)(r−5)!=(35−2r)(34−2r)(33−2r)(32−2r)(31−2r)(2r−1)!(r−5)!(39−r)(38−r)(37−r)(36−r)(35−r)=(2r−1)!(35−2r)(34−2r)(33−2r)(32−2r)(31−2r) At this point, we can compare the powers of the terms to solve for rr. Alternatively, we can notice that the factorials on both sides must be equal. Thus, we have: (39−r)(38−r)(37−r)(36−r)(35−r)=(35−2r)(34−2r)(33−2r)(32−2r)(31−2r)(39−r)(38−r)(37−r)(36−r)(35−r)=(35−2r)(34−2r)(33−2r)(32−2r)(31−2r) Solving this equation will give us the value of rr. However, this process involves some algebraic manipulation and arithmetic calculations which I can guide you through step by step during our tutoring session on UrbanPro. Feel free to reach out to me there, and we can work through this problem together! read less