How do we solve this C22011+ C52011 + C8 2011 +C100T2011 - Maths - Principle of Mathematical Induction

Question

How do we solve this

C 2 2011 +   C 5 2011   +   C 8   2011   +
C 100 T 2011

Zuhaib Ul Zamann · Accepted Answer

Dear student,
We consider the expansion of (1+x)2011=C0+C1x+
+C2011x2011Put x=1 we getC0+C1+C2+C3+
+C2011=22011     APut x=ωC0+C1ω+C2ω2+C3ω3+
+C2011ω2011=-ω4022=-ω2    BPut x=ω2 we getC0+C1ω2+C2ω4+C3ω6+
+C2011ω4022=-ω2011=-ω     CA+ωB+ω2C givesC01+ω+ω2+C11+ω2+ω+C21+1+1+C31+ω+ω2+C41+ω2+ω+C51+1+1+
+C20111+ω2+ω=22011-1-1⇒3C2+C5+C7+
C2009=22011-2⇒6C2+C5+ +C1007=22011-2⇒C2+C5+
+C1007=22010-13Regards

How do we solve this C 2 2011 + C 5 2011 + C 8 2011 . . . . + C 100 T 2011

How do we solve this

${}^{2011}C_{2} + {}^{2011}C_{5} + {}^{2011}C_{8} . . . . + {}^{2011}{\underset{100 T}{C}}$