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Given, sn=1+q+q^2++q^n ,Sn=1+(q+1)/2+((q...

Given, `s_n=1+q+q^2++q^n ,S_n=1+(q+1)/2+((q+1)/2)^2++((q+1)/2)^n ,q!=1` prove that `^n+1C_1+^(n+1)C_2s_1+^(n+1)C_3s_2++^(n+1)C_(n+1)s_n2^n S_ndot`

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`.^(n+1)C_1+.^(n+1)C_2s_1+.^(n+1)C_3s_2+...+.^(n+1)C_(n+1s_n)`
`=sum_(r=1)^(n+1).^(n+1)C_rs_(r-1)`,
where `s_n=1+q+q^2+...+q^n=(1-q^(n+1))/(1-q)`
` therefore sum_(r=1)^(n+1).^(n+1)C_r((1-q^r)/(1-q))=(1)/(1-q)(sum_(r=1)^(n+1).^(n+1)C_r-sum_(r=1)^(n+1).^(n+1)C_rq^r)`
` =(1)/(1-q)[(1+1)^(n+1)-(1+q)^(n+1)]`
` =(1)/(1-q)[2^(n+1)-(1+q)^(n+1)]....(i)`
Alos, `S_n=1((q+1)/(2))+((q+1)/(2))^2+...+((q+1)/(2))^n`
` =(1-((q+1)/(2))^(n+1))/(1-((q+1)/(2)))=(2^(n+1)-(q+1)^(n+1))/(2^n(1-q))....(ii)`
From Eqs. (i) and (ii),
` .^(n+1)C_1+ .^(n+1)C_2s_1+.^(n+1)C_3s_2+...+.^(n+1)C_n+1s_n=2.^nS_n`
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