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If .^(m)C(1)=.^(n)C(2) prove that m=(1)/...

If `.^(m)C_(1)=.^(n)C_(2)` prove that `m=(1)/(2)n(n-1)`.

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The value of .^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"….."+.^(n+m-1)C_(m) is equal to a. .^(m+n)C_(n) - 1 b. .^(m+n)C_(n-1) c. .^(m)C_(1) + ^(m+1)C_(2) + ^(m+2)C_(3) + "…." + ^(m+n-1)C_(n) d. .^(m+n)C_(m) - 1