Find the last two digits of the double summation above.
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∑ ∑ 0 ≤ i ≤ j ≤ 1 0 1 0 C j j C i
= ∑ j = 0 1 0 ∑ i = 0 j 1 0 C j j C i
As 1 0 C j is independent of variable i we can take it outside the inner sigma.
= ∑ j = 0 1 0 1 0 C j ∑ i = 0 j j C i
Using ∑ i = 0 j j C i = 2 j ,
= ∑ j = 0 1 0 1 0 C j 2 j
Using ( 1 + x ) 1 0 = ∑ r = 0 1 0 1 0 C r x r ,
= ∑ j = 0 1 0 1 0 C j 2 j = 3 1 0
= 3 1 0