The middle term in the expansion of (b√a5−5a√b)12 is:
12C6(b/a)3
-12C6(b/a)3
12C7(b5/a)
-12C7(b5/a)
Middle term is T122+1=T6+1=12C6(b√a5)12−6(−5a√b)6=12C6(ba)3
If a1,a2,a3,a4 are the coefficients of any four
consecutive terms in the expansion of (1+x)n, then
a1a1+a2 + a3a3+a4 =