For Tr+1 to be integer 6144−r‌=2k1 r‌=3k2 ⇒r should be multiple of 6 . ‌⇒r=0,6,12,18,...,6144 ‌⇒‌ Number of terms ‌=‌
6144
6
+1=1025 To find α1025−α1026−α1024 Let f(x)=(1+x)−1(1+x2)−1(1+x4)−1 ‌(1+x8)−1×(1+x16)−1 ‌=‌
1
(1+x)(1+x2)(1+x4)(1+x8)(1+x16)
‌=‌
1−x
1−x32
‌⇒(1−x)(1−x32)−1 ‌⇒(1−x32)−1−x×(1−x32)−1 ‌α1025=‌ Coefficient of ‌x1025‌ in ‌ ‌(1−x32)−1‌ - Coefficient of ‌x1024‌ in ‌ ‌(1−x32)−1 ‌α1025=−‌ Coefficient of ‌x1024‌ in ‌(1−x32)−1 ‌α1026=‌ Coefficient of ‌x1026‌ in ‌(1−x32)−1 ‌α1024=‌ Coefficient of ‌x1024‌ in ‌(1−x32)−1 So, α1025−α1026−α1024 =−2×‌ Coefficient of ‌x1024‌ in ‌(1−x32)−1 Coefficient of x1026 in (1−x32)−1 =−2×1−0=−2
