Prove that `.^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." + n xx .^(n)C_(n) = n2^(n-1)`. Hence, prove that `.^(n)C_(1).(.^(n)C_(2))^(2).(.^(n)C_(3))^(3)"......."(.^(n)C_(n))^(n) le ((2^(n))/(n+1))^(.^(n+1)C_(2)) AA n in N`.

Prove that `.^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." +

1.	Prove that `.^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." + n xx .^(n)C_(n) = n2^(n-1)`. Hence, prove that `.^(n)C_(1).(.^(n)C_(2))^(2).(.^(n)C_(3))^(3)"......."(.^(n)C_(n))^(n) le ((2^(n))/(n+1))^(.^(n+1)C_(2)) AA n in N`.
Answer» Method I : `.^(n)C_(1) + 2.^(n)C_(2)+3.^(n)C_(3)+"….."+n.^(n)C_(n)` `=underset(r=1)overset(n)sumr.^(n)C_(r)` `= n underset(r=1)overset(n)sum.^(n-1)C_(r-1)` `= n(.^(n-1)C_(0) + .^(n-1)C_(1) + .^(n-1)C_(2) + .^(n-1)C_(3)+"…."+.^(n-1)C_(n-1))` `= n2^(n-1)` Method II : We have `(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x+.^(n)C_(2)x^(2)+"...."+.^(n)C_(n)x^(n)` Differentiating w.r.t x, we get `n(1+x)^(n-1)= .^(n)C_(1)+2 xx .^(n)C_(2)x + 3 xx .^(n)C_(3)x^(2) +"......" + n xx .^(n)C_(n)x^(n-1)` Putting `x = 1`, we get `n2^( n-1) = .^(n)C_(1)+2xx .^(n)C_(2) + "....." + n xx .^(n)C_(n)` Using `A.M. ge G.M.`, we get `(.^(n)C_(1)+2.^(n)C_(2)+3.^(n)C_(3)+".....+n.^(n)C_(n))/((n(n+1))/(2))` `[(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)]^((1)/((pi(n-1))/2))` `rArr (n2^(n-1))/((n(n+1))/(2))ge [(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)]^((2)/(n(n+1)))` `(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)le((2^(n))/(n+1))^((n(n+1))/(2))` or `(.^(n)C_(1))(.^(n)C_(2))^(2)(.^(n)C_(3))^(3)"....."(.^(n)C_(n))^(n)le((2^(n))/(n+1))^(.^(n+1)C_(2))`

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Prove that `.^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." + n xx .^(n)C_(n) = n2^(n-1)`. Hence, prove that `.^(n)C_(1).(.^(n)C_(2))^(2).(.^(n)C_(3))^(3)"......."(.^(n)C_(n))^(n) le ((2^(n))/(n+1))^(.^(n+1)C_(2)) AA n in N`.

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