Hi friends,
I have this sum..∑10y=31y−2−∑10y=31y−1
What I did was:
For the first summation:
T1=1; T2=1/2...up to T10=1/8 = 761/280
For the second Summation
T1=1/2, T2=1/3...up to T10=1/9 = 1,828
Then just subtracting 1,828 from 761/280, and getting 0,89 as final answer..
Was my approach correct at all?..Thank you all for confirming/advising..I do appreciate.
HI Alan,
this is beautiful, thank you vwery much.
I do however have 2 questions please...
I would have mistakenly re-written the first term as 1,5+∑10k=31k
and then from here ∑10k=11k−1,5
This would have been completely wrong, and would not lead me in any way towards the end?
I have never seen the upper limit changed, as you have done?...Would ytou kindly just explain that to me, and just one thing more: The second term is re-written as 1k−1+19
I'm looking hard here but fail to see how you got to that?...Thank you for your time..
The upper limit can be written as y = 10, though the "y" is usually dropped for the upper limit.
With y - 2 = k what you do is ask what is the value of k when y = 3 (k = 3 - 2 = 1), and what is the value of k when y = 10 (k = 10 - 2 = 8).
∑k=9k=21k=∑k=9k=11k−11=∑k=8k=11k−11+19
The limits for the summations are the same so why not simply say
10∑y=31y−2−10∑y=31y−1=10∑y=3{1y−2−1y−1}=(1/1−1/2)+(1/2−1/3)+(1/3−1/4)+⋯+(1/8−1/9)=1−1/9=8/9.
Hi guest,
yes this was basically my approach, then Alan gave me a more elegant way, which I actually like, but just fail to see how he got certasin values...Thank you for giving your input as well, appreciated..