1.

यदि `(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+...+C_(n)x^(n)` तो साबित कीजिए कि `C_(0)-(C_(1))/(2)+(C_(2))/(3)-......+(-1)^(n)(C_(n))/(n+1)=(1)/(n+1)`

Answer» दिया गया श्रेणी है : `C_(0)-(C_(1))/(2)+(C_(2))/(3)-...+(-1)^(n).(C_(n))/(n+1)`
r वाँ पद, `t_(r)=(-1)^(r-1)(""^(n)C_(r-1))/(r)`
अब, `C_(0)-(C_(1))/(2)+(C_(2))/(3)-...+(-1)^(n).(C_(n))/(n+1)`
`=underset(r=1)overset(n+1)sum(-1)^(r-1).(""^(n)C_(r-1))/(r)=underset(r=1)overset(n+1)sum(-1)^(r-1).(""^(n+1)C_(r))/(n+1)" "[because (""^(n)C_(r-1))/(r)=(""^(n+1)C_(r))/(n+1)]`
`=(1)/(n+1)[""^(n+1)C_(1)-""^(n+1)C_(2)+""^(n+1)C_(3)-...+(-1)^(n).""^(n+1)C_(n+1)]`
`=(1)/(n+1)[-""^(n+1)C_(0)+""^(n+1)C_(1)-""^(n+1)C_(2).+""^(n+1)C_(3)-...+(-1)^(n).""^(n+1)C_(n+1)+""^(n+1)C_(0)]`
`=(1)/(n+1)[-{""^(n+1)C_(0)-""^(n+1)C_(1)+...+(-1)^(n+1).""^(n+1)C_(n+1)}+""^(n+1)C_(0)]`
`=(1)/(n+1)[-(1-1)^(n+1)+""^(n+1)C_(0)]=(1)/(n+1)`


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