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Question Number 195079 by sonukgindia last updated on 23/Jul/23

	Answered by witcher3 last updated on 24/Jul/23
	$we applie S_n to this sum witch symetric Groupe of n element we not this elements σ cardS_n =n!;{1.....n}→^σ {1,....,n} S=Σ_(k_1 +...+k_n =m) (1/(2^k_1 (2^k_1 +2^k_2 )....(2^k_1 +2^k_2 +...+2^k_n ))) =Σ_(k_1 +...+k_n =m) (1/(2^k_(σ(1)) (2^k_(σ(1)) +2^k_(σ(2)) )....(2^k_1 +2^k_2 +...+2^k_n ))) S=(1/(n!))Σ_(σ∈S_n ) Σ_(k_1 +...+k_n =m) (1/(2^k_(σ(1)) (2^(σ(k_1 )) +2^(σ(k_2 )) )....(2^k_1 +2^k_2 +...+2^k_n ))) Σ_(σ∈S_n ) (1/(2^(σ(k_1 )) (2^(σ(k_1 )) +2^(σ(k_2 )) ).....(2^k_1 +...+2^k_n )))=(1/2^(k_1 +k_2 +..+k_n ) ) proof tack τ(1,2) transposition of 1 &2 2S_1 =Σ(1/(2^k_1 .....(2^k_1 +..2^k_n )))+(1/(2^k_2 (2^k_1 +2^k_2 )...(2^k_2 +2^k_1 ....+2^k_n ))) S_2 =2S_1 =(1/(2^(k_1 +k_2 ) (2^k_1 +2^k_2 +2^k_3 )(2^k_1 +...+2^k_n ))) S_2 τ(1,3)+S_2 τ(2,3)+S_2 =S_3 =3S_2 =(1/(2^(k_1 +k_2 +k_3 ) (2^k_1 +2^k_2 +2^k_3 +2^k_4 )...(2^k_1 +2^k_2 +..2^k_n ))) S^((t)) =Σ_(k=1) ^(t−1) S_(m−1) τ(1,t)+S_(m−1) =(1/(2^(k_1 +...+k_t ) .(2^k_1 +2^k_2 +..+2^k_(t+1) )..(2^k_1 +2^k_2 +..+2^k_n ))) S^n =nS_(n−1) =(1/2^(k_1 +...+k_n ) ) n!S=Σ_(k_1 +..+k_n =m) (1/2^(k_1 +...+k_n ) )=(1/2^m ) Σ_(k_1 +..+k_n =m) 1=(1/2^m ) (((m+n−1)),(( m)) ) S=( ((( m+n−1)),(( m)) )/(n!2^m )) “Σ_(σ∈S_n ) (1/(2^k_(σ(1)) (2^(kσ(1)) +2^(kσ(2)) )...(2^(kσ(1)) +2^(kσ(2)) +2^(kσ(n)) ))=(1/2^(k_1 +k_2 +..+k_n ) )” was the key σ[1..n]=[1...n]⇒2^(kσ(1)) +...+2^(kσ(n)) =2^k_1 +...+2^k_n σ is bijection$
	$we applie S_{n} to this sum witch symetric Groupe of$ $n element we not this elements σ$ ${cardS}_{n} = n!; {1 . . . . . n} \overset{σ}{\to} {1, . . . ., n}$ $S = \sum_{k_{1} + . . . + k_{n} = m} \frac{1}{2^{k_{1}} (2^{k_{1}} + 2^{k_{2}}) . . . . (2^{k_{1}} + 2^{k_{2}} + . . . + 2^{k_{n}})}$ $= \sum_{k_{1} + . . . + k_{n} = m} \frac{1}{2^{k_{σ (1)}} (2^{k_{σ (1)}} + 2^{k_{σ (2)}}) . . . . (2^{k_{1}} + 2^{k_{2}} + . . . + 2^{k_{n}})}$ $S = \frac{1}{n!} \sum_{σ \in S_{n}} \sum_{k_{1} + . . . + k_{n} = m} \frac{1}{2^{k_{σ (1)}} (2^{σ (k_{1})} + 2^{σ (k_{2})}) . . . . (2^{k_{1}} + 2^{k_{2}} + . . . + 2^{k_{n}})}$ $\sum_{σ \in S_{n}} \frac{1}{2^{σ (k_{1})} (2^{σ (k_{1})} + 2^{σ (k_{2})}) . . . . . (2^{k_{1}} + . . . + 2^{k_{n}})} = \frac{1}{2^{k_{1} + k_{2} + . . + k_{n}}}$ $proof$ $tack τ (1, 2) transposition of 1 & 2$ ${2 S}_{1} = Σ \frac{1}{2^{k_{1}} . . . . . (2^{k_{1}} + . . 2^{k_{n}})} + \frac{1}{2^{k_{2}} (2^{k_{1}} + 2^{k_{2}}) . . . (2^{k_{2}} + 2^{k_{1}} . . . . + 2^{k_{n}})}$ $S_{2} = {2 S}_{1} = \frac{1}{2^{k_{1} + k_{2}} (2^{k_{1}} + 2^{k_{2}} + 2^{k_{3}}) (2^{k_{1}} + . . . + 2^{k_{n}})}$ $S_{2} τ (1, 3) + S_{2} τ (2, 3) + S_{2} = S_{3} = {3 S}_{2} = \frac{1}{2^{k_{1} + k_{2} + k_{3}} (2^{k_{1}} + 2^{k_{2}} + 2^{k_{3}} + 2^{k_{4}}) . . . (2^{k_{1}} + 2^{k_{2}} + . . 2^{k_{n}})}$ $S^{(t)} = \underset{k = 1}{\sum^{t - 1}} S_{m - 1} τ (1, t) + S_{m - 1} = \frac{1}{2^{k_{1} + . . . + k_{t}} . (2^{k_{1}} + 2^{k_{2}} + . . + 2^{k_{t + 1}}) . . (2^{k_{1}} + 2^{k_{2}} + . . + 2^{k_{n}})}$ $S^{n} = {nS}_{n - 1} = \frac{1}{2^{k_{1} + . . . + k_{n}}}$ $n! S = \sum_{k_{1} + . . + k_{n} = m} \frac{1}{2^{k_{1} + . . . + k_{n}}} = \frac{1}{2^{m}} \sum_{k_{1} + . . + k_{n} = m} 1 = \frac{1}{2^{m}} (\begin{matrix} m + n - 1 \\ m \end{matrix})$ $S = \frac{(\begin{matrix} m + n - 1 \\ m \end{matrix})}{n! 2^{m}}$ $‘ ‘ \sum_{σ \in S_{n}} \frac{1}{2^{k_{σ (1)}} (2^{k σ (1)} + 2^{k σ (2)}) . . . (2^{k σ (1)} + 2^{k σ (2)} + 2^{k σ (n)}} = {\frac{1}{2^{k_{1} + k_{2} + . . + k_{n}}}}^{″}$ $was the key$ $σ [1 . . n] = [1 . . . n] \Rightarrow 2^{k σ (1)} + . . . + 2^{k σ (n)} = 2^{k_{1}} + . . . + 2^{k_{n}} σ is bijection$
	Commented by witcher3 last updated on 24/Jul/23

	$car (k_{1}, k_{2} . . . k_{n}) ∣ (k_{1} + . . . + k_{n} = m) = (\begin{matrix} m + n - 1 \\ m \end{matrix})$
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