C=1/21.22.23+1/22.23.24++1/200.201. 202

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It looks like you have to find the value of the sum,

so that the n-th term in the sum is

for 1 ≤ n ≤ 180.

We can then write the sum as

Break up the summand into partial fractions:

Combine the fractions into one with a common denominator and set the numerators equal to one another:

Expand the right side and collect terms with the same power of n :

Then

a + b + c = 0

43a + 42b + 41c = 0

462a + 440b + 420c = 1

==>   a = 1/2, b = -1, c = 1/2

Now our sum is

which is a telescoping sum. If we write out the first and last few terms, we have

C = 1/(2×21) - 1/22 + 1/(2×23)

… … + 1/(2×22) - 1/23 + 1/(2×24)

… … + 1/(2×23) - 1/24 + 1/(2×25)

… … + 1/(2×24) - 1/25 + 1/(2×26)

… … + … - … + …

… … + 1/(2×198) - 1/199 + 1/(2×200)

… … + 1/(2×199) - 1/200 + 1/(2×201)

… … + 1/(2×200) - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

C = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

C = 1/42 - 1/44 - 1/402 + 1/404

C = 1,115/1,042,118 ≈ 0.00106994

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day