find the value of : (1/(3^2 -2^2)) + (1/ (7^2 -2^2 )) + (1/ (11^2 -2^2)) +...........+ (1/(39^2-2^2))?

Question

how to solve this question.
ans given in book is 10/41

Answer 1

S = Sum(1/((4i-1)^2-2^2)) from i=1 to i=10

Notice that the denominators are all differences of squares:
S = Sum(1/(4i-1-2) - 1/(4i-1+2))/4 from i=1 to i=10

Then simplify:
S = Sum(1/(4i-3) - 1/(4i+1))/4 from i=1 to 10

The intermediate terms all cancel out leaving only the first and last term:
S = (1 - 1/41)/4 = (40/41)/4 = 10/41

Answer 2

1/(9-4) + 1/(49-4) + 1/(121-4) + 1/(225-4) + 1/(361-4) + 1/(529-4) + 1/(729-4) + 1/(961-4) + 1/(1225-4) + 1/(1521-4)

=

1/5 + 1/45 + 1/117 + 1/221 + 1/357 + 1/525 + 1/725 + 1/957 + 1/1221 + 1/1517

(the least common multiple of these denominators is: 168441498225, so my computer tells me)

= 33688299645/168441498225 + 3743144405/168441498225 + 1439670925168441498225 + 762178725/168441498225 + 471824925/168441498225 + 320840949/168441498225 + 232333101/168441498225 + 176009925168441498225 + 137953725/168441498225 + 111035925/168441498225

= 41083292250/168441498225

(To reduce, divide both top and bottom by 4108329225)

= 10/41

Answer 3

t1 = 1/(3 - 2)(3 + 2) = 1/4(1/(3 - 2) - 1/(3 + 2)) = 1/4(1/1 - 1/5)
t2 = 1/(7 - 2)(7 + 2) = 1/4(1/(7 - 2) - 1/(7 + 2)) = 1/4(1/5 - 1/9)
t3 =1/(11 -2)(11+2)=1/4(1/(11 - 2) -1/(11 + 2))= 1/4(1/9 - 1/13)
...
...
t9 =1/(39 -2)(39+2)=1/4(1/(39 - 2)-1/(39 + 2))=1/4(1/37 -1/41)


Adding vertically we get req value = 1/4(1 - 1/41) = 10/41

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