Solution
1. Find the General Term
The general term Tk+1 in the binomial expansion of (a+b)n is given by Tk+1=nCk⋅an−k⋅bk.
For the expression (2x51−x−51)15:
Tk+1=15Ck⋅(2x51)15−k⋅(−x−51)k
Tk+1=15Ck⋅215−k⋅(−1)k⋅x515−k⋅x−5k
Tk+1=15Ck⋅215−k⋅(−1)k⋅x515−2k
2. Find the Coefficient m (for x−1)
To find the coefficient of x−1, we set the exponent of x to −1:
515−2k=−1⟹15−2k=−5⟹2k=20⟹k=10
Substituting k=10 into the general term:
m=15C10⋅215−10⋅(−1)10=15C10⋅25
3. Find the Coefficient n (for x−3)
To find the coefficient of x−3, we set the exponent of x to −3:
515−2k=−3⟹15−2k=−15⟹2k=30⟹k=15
Substituting k=15 into the general term:
n=15C15⋅215−15⋅(−1)15=1⋅20⋅(−1)=−1
4. Solve for r
We are given the condition mn2=15Cr⋅2r.
Substitute the values of m and n:
(15C10⋅25)⋅(−1)2=15Cr⋅2r
By comparing both sides:
Since 15C10=15C5, the equation holds true for r=5.
Final Answer:
The value of r is 5.
