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Prove that ""^(n)C_(3)+""^(n)C_(7) + ""^(n)C_(11) + ...= 1/2{2^(n-1) - 2^(n//2 )sin"" (npi)/(4)} |
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Answer» Solution :In given series difference in lower suffices is 4. i.e., ` 7 -3 =11 = …= 4 ` Now , ` (1)^(1//4) = (cos 0+ I sin 0 )^(1//4)` `= (cos 2r pi + i sin2r pi)^(1//4)` `= cos ""(rpi)/(2) + i sin"" (rpi)/(4) "where r" = 0,1,2,3 ` FOUR roots of unity` = 1,i,-1, 1,alpha, alpha^(2) , alpha^(3)` [SAY] and `(1 + x)^(n) = sum_(r=0)^(n) ""^(n)C_(r) x^(r)` Putting ` x = 1, alpha, alpha^(2) , alpha^(3) , " we GET " 2^(n) = sum_(r=0)^(n) ""^(n)C_(r)` ...(i) ` (1 + alpha)^(n) = sum_(r=0)^(n) ""^(n)C_(r) alpha^(r)` ...(ii) ` (1 + alpha ^(2))^(n) = sum_(r=0)^(n) ""^(n)C_(r) alpha^(2r) `...(iii) `(1 + alpha^(2))^(n) - sum_(r=0)^(n) ""^(n)C_(r) alpha^(3r)` ...(iv) On multiplying Eq.(i) by 1, Eq.(ii) By ` alpha ` , Eq.(iii) by ` alpha^(2)` and Eq.(iv) by ` alpha^(3)` and adding , we get ` rArr 2^(n) + alpha (1 + alpha)^(n) + alpha^(2) (1 + alpha^(2))^(n) + alpha^(3) (1 + alpha^(3))^(n)` ` =sum_(r=0)^(n) ""^(n)C_(r) (1 + alpha ^(r+1) + alpha^(2r +2)+ alpha^(3r +3))` ...(v) for r = 3,7,11,... RHS of Eq.(v) ` =""^(n)C_(r) (1 + alpha ^(4) + alpha^(8)+ alpha^(12))+""^(n)C_(7) (1 + alpha ^(4) + alpha^(16)+ alpha^(24))+""^(n)C_(11) (+ alpha ^(12) + alpha^(24)+ alpha^(36))+ ... ` ` = 4(""^(n)C_(3) + ""^(n)C_(7) + ""^(n)C_(11) +...) "" [ because alpha^(4) = 1]` and LHS of Eq.(v) `= 2^(n) + ii (1 + i)^(n) + i^(2) (1 + i^(2))^(n) + i^(3) (1 + i^(3))^(n)` ` = 2^(n) + i (1+ i)^(n) + 0 - i (1 - i)^(n)` ` = 2^(n) + i {(1 + i)^(n) - (1 - i)^(n)}` Since , `[(1 + i)^(n)= [sqrt(2)((1)/(sqrt(2))+ (i)/(sqrt(2)))]^(n)]` ` = 2^(n) + i2^(n//2)* sin"" (npi)/(4) = 2^(n//2) {cos""(pi)/(4) + i sin"" (pi)/(4)}^(n)` ` = 2^(n) - i2^(n//2)* sin"" (npi)/(4) = 2^(n//2) {cos""(pi)/(4) + i sin"" (pi)/(4)}` HENCE , ` 4(""^(n)C_(3) + ""^(n)C_(7) + ""^(n)C_(11)+ ...) = 2 (2^(n-1) - 2^(n//2) sin" " (npi)/(4))` ` rArr ""^(n)C_(3) + ""^(n)C_(7) + ""^(n)C_(11) + ... = 1/2 (2^(n-1) - 2^(n//2) sin"" (npi)/(4))`. |
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