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 – 6 = 34C2r – 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