juriemagic

Thank you Alan....I'll go through this again carefully...much appreciated

Thank you

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..

Guest,

please, how do you get a=3/2?

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 \(1,5+\sum_{k=3}^{10}{1 \over k}\)

and then from here \(\sum_{k=1}^{10}{1 \over k}-1,5\)

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 \({1 \over k}-1+{1 \over 9}\)

I'm looking hard here but fail to see how you got to that?...Thank you for your time..

Understood..thank you Alan. Stay blessed..

Hi Alan,

Gosh, what a piece of work!!..Thank you very much indeed.

I follow the sum up to the point where \(2j +3\) changes to \(j+3(n-2)\),

Could I ask if you would please just explain that bit to me?...Thanks Alan...

Hi friends,

I tried this one again this morning from a different angle..still struggle with it, I promise after this one there are no more of these...please help..

I got y=1, and also x=1..

Hello Melody,

You know...I am out of words here...I really just do not think...honestly...Starting with 3, ending with 4...obviously the two terms must be added....and that equals 5y....I tend to make things way more difficult than they really are....Thank you Melody..appreciate