If an= summation 1/nCr then summation r/nCr is equal to (a)(n-1)an (b)nan (c)1/2nan (d)NOT | BT PYQ🔥

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If `a_n=sum(r=0)^n 1/^nC_r, then sum _(r=0^n r/^nC_r` equals (A) `(n-1)a_n` (B) `na_n` (C) `1/2na_n` If `a_n=sum(r=0)^n 1/^nC_r, then sum _(r=0^n r/^nC_r` equals (A) `(n-1)a_n` (B) `na_n` (C) `1/2na_n`
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