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+1)C_1+^(n+1)C_2s_1+^(n+1)C_3s_2++^(n+1)C_(n+1)s_n=2^n S_ndot`

Explanation

`s_(n)` is in geometric progression, hence,
`s_(n) = (q^(n+1) -1)/(q-1) . q ne 1`
Also, `S_( n)` is geometric progression.
`:. S_(n) = (((q+1)/(2))^(n+1)-1)/(((q+1)/(2))-1) = ((q+1)^(n+1).-2^(n+1))/(2^(n)(q-1))`
Consider `.^((n+1))C_(1)+.^((n+1))C_(2)s_(1)+.^((n+1))C_(3)s_(3)+"....."+.^((n+1))C_(n+1)s_(n)`
`.^((n+1))C_(1)((q-1)/(q-1))+.^((n+1))C_(2)(q^(2)-1)(q-1)+"......"+.^((n+1))C_(n+1)(q^(n+1)-1)/(q-1)`
`=(1/(q-1))[{.^((n+1))C_(1)q+.^((n+1))C_(2)q^(2)+"...."+.^((n+1))C_(n+1)q^(n+1)}-{.^((n+1))C_(1)+.^((n+1))C_(2)+"....."+.^((n+1))C_(n+1)}]`
`= ((1)/(q-1)) [{(1+q)^(n+1)-1}-{2^(n+1)-1}]`
Since
`{{:((1+q)^(n+1)=.^(n+1)C_(0)+.^((n+1))C_(1)q+.^((n+1))C_(2)q^(2)+"...."+.^((n+1))C_(n+1)q^(n+1)),(" "2^(n-1)=.^((n+1))C_(0)+.^((n+1))C_(1)+"...."+.^((n+1))C_(n+1)):}}`
`rArr ((1)/(q-1))[(1+q)^(n+1)-2^(n+1)-2^(n+1)]= ((1+q)^(n+1)-2^(n+1))/(q-1)`
Thus, `.^((n+1))C_(1)+.^((n+1))C_(2)s_(1)+"....."+.^((n+1))C_(n+1)s_(n) = ((1+q)^(n+1)-2^(n+1))/(q-1)`
Therefore, `.^(n+1)C_(1) + .^((n+1))C_(2)s_(1) + "....."+.^((n+1))C_(n+1)s_(n+1)=2^(n)S_(n)` (Using (1))