If _r=0^10 0.05em ( 10^r+1-110^r ) ^11C_r+1= 11-11^1110^10, then ,is equal to:

Question

Accepted Answer

Correct answer: B. (b) _r=0^10 0.05em ( 10^r-1-110^r ) ^11C_r+1 = _r=0^10 0.05em(10- 110^r) ^11C_r+1 =10 _r=0 0.05em ^11C_r+1-10 ( ^11C_r+1 ( 110 )^r+1 ) =10( ^11C_1+ ^11C_2+ … . .+ ^11C_11) =10( ^11C_1 ( 110 )^1+ ^11C_2 ( 110 )^2+ … .+ ^11C_11 ( 110 )^11) =10(2^11-1)-10((1+ 110)^11-1) =10 (2 )^11-10- 11^1110^10+10 = (20 )^11-11^1110^10 ∴ =20

$If \sum_{r = 0}^{10} (\frac{10 ^{r + 1} - 1}{10 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α 11 - 11 ^{11}}{10 ^{10}}, then α,is equal to:$

Solution

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