If ∑r=010(10r10r+1−1)⋅11Cr+1=1010α11−1111, then α is equal to:
Get the complete, step-by-step math solution for: "If _{r=0}^{10} {10^{r+1} - 1}{10^r} · {}^{11}C_{r+1} = {α^{11} - 11^{11}}{10^{10}}, then α is equal to:". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Simplify the general term of the summation
First, we simplify the expression inside the summation. We can split the fraction 10r10r+1−1 into two terms: 10r10r+1−10r1. This simplifies to 10−10r1.
Step 2: Substitute the simplified term into the summation
Now we substitute the simplified term back into the summation. We can then split the summation into two separate summations due to the distributive property.
Step 3: Adjust the index of summation for the first term
For the first summation, let k=r+1. When r=0, k=1. When r=10, k=11. So the summation becomes 10∑k=11111Ck. We know that ∑k=0nnCk=2n, so ∑k=1nnCk=2n−nC0=2n−1. Therefore, 10∑k=11111Ck=10(211−1).
Step 4: Adjust the index of summation for the second term
For the second summation, again let k=r+1. The summation becomes ∑k=11110k−1111Ck. We can rewrite 10k−11 as 10⋅10k1. So the term is 10∑k=11111Ck(101)k.
Step 5: Apply the binomial theorem to the second term
Recall the binomial theorem: (a+b)n=∑k=0nnCkan−kbk. Here, we have a=1, b=101, and n=11. So, ∑k=01111Ck(101)k=(1+101)11. Since our summation starts from k=1, we subtract the k=0 term, which is 11C0(101)0=1⋅1=1. Thus, the second term simplifies to 10[(1011)11−1].
Step 6: Combine the two terms and solve for α
Now we combine the simplified first and second terms. 10(211−1)−10[(1011)11−1]=10⋅211−10−10⋅10111111+10. The −10 and +10 cancel out. The term 10⋅10111111 simplifies to 10101111. So the entire expression is 10⋅211−10101111. We are given that this is equal to 1010α11−1111. We can rewrite 10⋅211 as 10101011⋅211=1010(10⋅2)11=10102011. Therefore, 10102011−1111=1010α11−1111. Comparing the numerators, we get α11=2011, which implies α=20.