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1 + 2 1 9 4 7 1 + 2 1 9 4 7 + 2 1 9 4 7 1
= 1 + 2 1 9 4 7 1 + 1 + 2 1 9 4 7 1 ( 2 1 9 4 7 1 )
= ( 1 + 2 1 9 4 7 1 ) ( 1 + 2 1 9 4 7 1 )
= ( 2 1 9 4 7 2 1 9 4 7 + 1 ) ( 1 + 2 1 9 4 7 1 )
= 2 1 9 4 7 1
Similarly pairing remaining terms (Note that there is no term without a pair), we get,
∑ n = 0 1 9 4 7 2 n + 2 1 9 4 7 1
= 2 1 9 4 7 9 7 4
This is what I did in the exam but didn't mark the option as there was no option matching to the above answer. And I got home and realised what a fool I was, FORGOT to cancel common factor 2. AGHHH!!!!!!!!
So, the answer is
= 2 1 9 4 5 4 8 7
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