Question 1205325
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Tutor @ikleyn probably has the most efficient pathway, but here's another approach.


The pattern in the numerator (1^2,3^2,5^2,7^2,...) follows the sequence (2k-1)^2 where k is an integer and k = 1 is the starting index.
Note how 1,3,5,7,... is an arithmetic sequence.
The highest that k goes is k = 52 because 2k-1 = 2*52 - 1 = 103.


The pattern in the denominator (2,6,10,14,...) is also arithmetic and follows the sequence 4k-2
Note that 4k-2 = 4*52 - 2 = 206.


Each term can be written of the form {{{((2k-1)^2)/(4k-2) =((2k-1)^2)/(2(2k-1)) = (2k-1)/2 = k - 1/2}}}


We're tasked to find the summation {{{sum((k-1/2),k=1,52)}}}


Use these summation identities
{{{sum((A+B),k=1,n) = (sum(A,k=1,n))+(sum(B,k=1,n))}}}
and
{{{sum(k,k=1,n) = n(n+1)/2}}}
and
{{{matrix(1,6,sum(c,k=1,n) = cn,"where","c","is","a","constant")}}} 
to have the following steps


{{{sum((k-1/2),k=1,52)=(sum(k,k=1,52))+(sum(-1/2,k=1,52))}}}


{{{sum((k-1/2),k=1,52)=52*(52+1)/2+(-1/2)*52}}}


{{{sum((k-1/2),k=1,52)=1352}}}


Therefore,
{{{matrix(1,5,(1^2)/2+(3^2)/6+(5^2)/10+(7^2)/14,"+","...","+",(103^2)/206) = 1352}}}
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