summation of series (method of differences): when we use ∑ f(r)- f(r-1),then it will

[cai cen ho]

summation of series (method of differences): when we use ∑ f(r)- f(r-1),then it will

according to the book example given,
when we use
∑ f(r)- f(r-1),then it will be
f(n) - f(0)...my question is why not it be like
f(1) -f(0)

Also,we use ∑f(r)- f(r+1)then, it ll be
f(1)-f(n+1),. why not it be f(n)-f(1+1)?.
did anyone know that?

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[Deleted member 4993]

according to the book example given,
when we use
∑ f(r)- f(r-1),then it will be
f(n) - f(0)...my question is why not it be like
f(1) -f(0)

Also,we use ∑f(r)- f(r+1)then, it ll be
f(1)-f(n+1),. why not it be f(n)-f(1+1)?.
did anyone know that?

Sent from my Lenovo A5000 using Tapatalk

Is your assignment:

∑ [f(r)- f(r-1)] = Sum

If it is, then those parentheses [] are super important. Without those, the meaning of the problem changes. If the problem is as I have presented then:


Sum = ∑[f(r)- f(r-1)] = [f(1) - f(0)] + [f(2) - f(1)] + [f(3) - f(2)] .... = -f(0) + f(3) + ......

What will happen if I expand above to 'n' terms....