If 26(233(C212)+255(C412)+277(C612)+...+21313(C1212))=313–α, then α is equal to: [2026]
(3)
26[233(C212)+255(C412)+277(C612)+...+21313(C1212)]=313–α
=2[23·C313+25·C513+27·C713+...+213·C1313]
=2[(1+2)13–(1–2)132–26]
=313+1–52=313–51 ∴ α=51