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Question Number 53141 by gunawan last updated on 18/Jan/19

If n is an integer greater than unity, then the value of a−^n C_1 (a−1)+^n C_2 (a−2)−^n C_3 (a−3)+... ..+(−1)^n (a−n) is

$$\mathrm{If}\:{n}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{unity},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${a}−\:^{{n}} {C}_{\mathrm{1}} \left({a}−\mathrm{1}\right)+\:^{{n}} {C}_{\mathrm{2}} \left({a}−\mathrm{2}\right)−\:^{{n}} {C}_{\mathrm{3}} \left({a}−\mathrm{3}\right)+... \\ $$$$\:\:\:\:\:\:\:..+\left(−\mathrm{1}\right)^{{n}} \left({a}−{n}\right)\:\mathrm{is} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19

a(1−nc_1 +nc_2 −nc_3 +...+(−1)^n nc_n )+(0+nc_1 −2nc_2 +3nc_3 −nc_4 +..−(−1)^n n) as_1 +s_2 (1+x)^n =1+nc_1 x+nc_2 x^2 +nc_3 x^3 +...+nc_n x^n put x=−1 0=1−nc_1 +nc_2 −nc_3 +..+(−1)^n nc_n so s_1 =0 n(1+x)^(n−1) =0+nc_1 +2nc_2 x+3nc_3 x^2 +...+n×nc_n x^(n−1) put x=−1 n×0=0+nc_1 −2nc_2 +3nc_3 −...+(−1)^n ×n×nc_n so s_2 =0 hence as_1 +s_2 =a×0+0=0

$${a}\left(\mathrm{1}−{nc}_{\mathrm{1}} +{nc}_{\mathrm{2}} −{nc}_{\mathrm{3}} +...+\left(−\mathrm{1}\right)^{{n}} {nc}_{{n}} \right)+\left(\mathrm{0}+{nc}_{\mathrm{1}} −\mathrm{2}{nc}_{\mathrm{2}} +\mathrm{3}{nc}_{\mathrm{3}} −{nc}_{\mathrm{4}} +..−\left(−\mathrm{1}\right)^{{n}} {n}\right) \\ $$$${as}_{\mathrm{1}} +{s}_{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nc}_{\mathrm{1}} {x}+{nc}_{\mathrm{2}} {x}^{\mathrm{2}} +{nc}_{\mathrm{3}} {x}^{\mathrm{3}} +...+{nc}_{{n}} {x}^{{n}} \\ $$$${put}\:{x}=−\mathrm{1} \\ $$$$\mathrm{0}=\mathrm{1}−{nc}_{\mathrm{1}} +{nc}_{\mathrm{2}} −{nc}_{\mathrm{3}} +..+\left(−\mathrm{1}\right)^{{n}} {nc}_{{n}} \\ $$$$\boldsymbol{{so}}\:\:\boldsymbol{{s}}_{\mathrm{1}} =\mathrm{0} \\ $$$$\boldsymbol{{n}}\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\boldsymbol{{n}}−\mathrm{1}} =\mathrm{0}+\boldsymbol{{nc}}_{\mathrm{1}} +\mathrm{2}\boldsymbol{{nc}}_{\mathrm{2}} \boldsymbol{{x}}+\mathrm{3}\boldsymbol{{nc}}_{\mathrm{3}} \boldsymbol{{x}}^{\mathrm{2}} +...+\boldsymbol{{n}}×\boldsymbol{{nc}}_{\boldsymbol{{n}}} \boldsymbol{{x}}^{\boldsymbol{{n}}−\mathrm{1}} \\ $$$$\boldsymbol{{put}}\:\boldsymbol{{x}}=−\mathrm{1} \\ $$$$\boldsymbol{{n}}×\mathrm{0}=\mathrm{0}+\boldsymbol{{nc}}_{\mathrm{1}} −\mathrm{2}\boldsymbol{{nc}}_{\mathrm{2}} +\mathrm{3}\boldsymbol{{nc}}_{\mathrm{3}} −...+\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} ×\boldsymbol{{n}}×\boldsymbol{{nc}}_{\boldsymbol{{n}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{s}}_{\mathrm{2}} =\mathrm{0} \\ $$$$\boldsymbol{{hence}}\:\boldsymbol{{as}}_{\mathrm{1}} +\boldsymbol{{s}}_{\mathrm{2}} \\ $$$$=\boldsymbol{{a}}×\mathrm{0}+\mathrm{0}=\mathrm{0} \\ $$

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