Question

proof me rongh

let an AP S-n =1+2+.........+n
and S-(n+1) = 0+11+1+........+(n+1)
if S-n = n(2a+(n-1)d)/2

and S_(n+1) = (n+1)(2a + (n)d)/2
where a is first term d is common difference

1/(S_n) - 1/(S_(n+1) = 2/n (2a +(n-1)d) - 2/(2a-(n)d) ........(i)


here by putting the value of a and d we can simplify (i)

1/(S_n) - 1/(S_(n+1) =2{ 2(2n+3)/(n^2 +2n - 2)(n^2 +4n +1) .................(1') or can be said to be
new eq(1)

or i think differently

why dont we just take S_n = (n^2 + 2n -2)/2 and S_(n+1) =(n^2 +4n+1)/2
and
when we takes the roots of these we get S_n =(n+1+i)(n-i+1)/2 and S_(n+1) = (n+2-(\sqrt{3} )(n+\sqrt{3} --2)/2

we get 1/S_n -1/S_(n+1) = 2/(n+i+1)(n-i+1) - 2/(n+2-(\sqrt{3} )(n+\sqrt{3} --2) ..........(2)

now putting n=1 eq(1') is not equal to eq(2)

PROOF ME RONGH (or find an error )

Answer 1

Answer:

whats the question................

Answer 2

Answer:

No there is no error in your solution.

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