If `n >2,` then prove that `C_1(a-1)-C_2xx(a-2)++(-1)^(n-1)C_n(a-n)=a ,w h e r eC_r=^n C_rdot`
If `n >2,` then prove that `C_1(a-1)-C_2xx(a-2)++(-1)^(n-1)C_n(a-n)=a ,w h e r eC_r=^n C_rdot`
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`S = C_(1)(a-1)- C_(2)(a-2) + "…." + (-1)^(n-1)C_(n)(a-n)`
`:. T_(r) = (-1)^(r-1)(a-r).^(n)C_(r)`
`= (-1)^(r-1)(a.^(n)C_(r) - r.^(n)C_(r))`
`= (-1)^(r-1)(a.^(n)C_(r)-n.^(n-1)C_(r-1))`
` = - a (-1)^(r ). .^(n)C_(r) - n (-1)^(r=1 xx n - 1) C_(r-1)`
Now, `S = underset(r=1)overset(n)sumT_(r)`
`= -a[(1-1)^(n)-.^(n)C_(0)] - n(1-1)^(n-1)`
`= an`
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