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Biren Omar Deshpande
Asked: 2022-11-11T09:10:15+05:30 In:

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)`

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)`
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  1. 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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Uma Dhiraj Meda
Asked: 2022-11-08T09:15:51+05:30 In:

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

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

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Jagruti Sohail Palla
Asked: 2022-11-07T23:46:31+05:30 In:

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)+…+(C_(n))/(n+1)=(2^(n+1)-1)/(n+1)” or “sum+(k=0)^(n)(C_(k))/(k+1)=(2^(n+1)-1)/(n+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)+…+(C_(n))/(n+1)=(2^(n+1)-1)/(n+1)” or “sum+(k=0)^(n)(C_(k))/(k+1)=(2^(n+1)-1)/(n+1)`
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  1. दिया गया श्रेणी है : `C_(0)+(C_(1))/(2)+(C_(2))/(3)+…+(C_(n))/(n+1)`
    r वाँ पद `t_(r)=(“”^(n)C_(r-1))/(r)`
    अब, `C_(0)+(C_(1))/(2)+(C_(2))/(3)+…+(C_(n))/(n+1)`
    `=underset(r=1)overset(n+1)sum(“”^(n)C_(r-1))/(r)=underset(r=1)overset(n+1)sum(“”^(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_(n+1))`
    `=(1)/(n+1)(“”^(n+1)C_(0)+””^(n+1)C_(1)+””^(n+1)C_(2)+…+””^(n+1)C_(n+1)-“”^(n+1)C_(0))`
    `=(1)/(n+1)(2^(n+1)-1)=(2^(n+1)-1)/(n+1)” “[because””^(n+1)C_(0)=1]`

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