Solve: 1/1*2*3 + 1/2*3*4 +... 1/8*9*10
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I'm getting 11/45... Multiply and divide by two at start.
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Divyankasc:
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In these summations , we first write the nth term
Notice here the nth term will be 1/(n)(n+1)1/(n)(n+1)
It can be written as (n+1)−n/(n)(n+1)(n+1)−n/(n)(n+1)
=>1/n−1/(n+1)=>1/n−1/(n+1)
Now when you take summations from n=1 to infinity, you can observe cancellation of terms
T1 =1/1−1/2=1/1−1/2
T2 =1/2−1/3=1/2−1/3
T3 =1/3−1/4=1/3−1/4
And so on till infinity
As by calculation we will be left with 1 - 1/infinity1/infinity the negative term will be approximately zero therefore the answer will be 1
Hope this helps!
Notice here the nth term will be 1/(n)(n+1)1/(n)(n+1)
It can be written as (n+1)−n/(n)(n+1)(n+1)−n/(n)(n+1)
=>1/n−1/(n+1)=>1/n−1/(n+1)
Now when you take summations from n=1 to infinity, you can observe cancellation of terms
T1 =1/1−1/2=1/1−1/2
T2 =1/2−1/3=1/2−1/3
T3 =1/3−1/4=1/3−1/4
And so on till infinity
As by calculation we will be left with 1 - 1/infinity1/infinity the negative term will be approximately zero therefore the answer will be 1
Hope this helps!
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