If 7=5+71(5+α)+771(5+2α)+771(5+3α)+∞, then the value of α is:
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Step-by-Step Solution
Step 1: Rearrange the equation
First, we isolate the infinite series by subtracting 5 from both sides of the given equation. This simplifies the problem to finding the value of α that makes the infinite series equal to 2.
Step 2: Split the series into two geometric progressions
We can split the given series into two separate series: one involving the constant term 5 and another involving α. This makes it easier to identify the patterns and sum each series individually.
Step 3: Identify the first series as a geometric progression
The first series is S1=71+771+7771+…. This is a geometric progression with first term a=71 and common ratio r=111. The sum of an infinite geometric progression is given by S=1−ra when ∣r∣<1.
Step 4: Calculate the sum of the first series
Using the formula for the sum of an infinite geometric progression, we substitute a=71 and r=111 to find the sum of the first series.
Step 5: Identify the second series as an arithmetico-geometric progression
The second series is S2=71+772+7773+…. This is an arithmetico-geometric progression where the arithmetic part is 1,2,3,… and the geometric part is 71,771,7771,…. The sum of an infinite arithmetico-geometric progression S=a+(a+d)r+(a+2d)r2+… is given by S=1−ra+(1−r)2dr. Here, a=1, d=1, and r=111.
Step 6: Calculate the sum of the second series
We substitute the values a=1, d=1, and r=111 into the formula for the sum of an infinite arithmetico-geometric progression to find S2.
Step 7: Substitute sums back into the rearranged equation and solve for α
Now we substitute the calculated sums S1=7011 and S2=100121 back into the equation from Step 2. Then, we solve the resulting linear equation for α.