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 rth  in the expansion of (1 + x)n  is nCr – 1.

 

∴ Coefficients of (r – 5)th and (2r – 1)th term in the Expansion of (1 + x)34 are 34Cr – 6 and 34C2r – 2 

 

It is given that these coefficients are equal

34Cr – 634C2r – 2 

r – 6 = 2r – 2 or r – 6 + 2r – 2 = 34                   (nCr = nCS ⇒ r = S or r + S = n)

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

⇒ 3r  = 34 + 8 = 42

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thanx sir

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