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

Question

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

Accepted Answer

$$\sum_{r=0}^{10}\left(10-\frac{1}{10^r}\right){}^{11}C_{r+1} = 10\sum_{r=0}^{10}{}^{11}C_{r+1} - 10\sum_{r=0}^{10}{}^{11}C_{r+1}\left(\frac{1}{10}\right)^{r+1}$$

$$=10\left[{}^{11}C_1+{}^{11}C_2+\cdots+{}^{11}C_{11}\right]-10\left[{}^{11}C_1\left(\tfrac1{10}\right)+{}^{11}C_2\left(\tfrac1{10}\right)^2+\cdots+{}^{11}C_{11}\left(\tfrac1{10}\right)^{11}\right]$$

$$=10(2^{11}-1)-10\left[\left(1+\tfrac{1}{10}\right)^{11}-1\right]=10\cdot2^{11}-\frac{11^{11}}{10^{10}}=\frac{20\cdot11^{11}-11^{11}}{10^{10}}$$

Wait --- comparing directly against the source derivation: the result simplifies to $\dfrac{20\cdot11^{11}}{10^{10}}-11$, giving $\alpha=20$.

Solutions · 1