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Question:

LetSk, k=1,2, …,100, denote the sum of the infinite geometric series whose first term isk𕒵k!and the common ratio is1k.Then, the value of1002100!+100∑k=2|(k2𕒷k+1)Sk|is :1243LetSk, k=1,2, …,100, denote the sum of the infinite geometric series whose first term isk𕒵k!and the common ratio is1k.Then, the value of1002100!+100∑k=2|(k2𕒷k+1)Sk|is :1243Sk, k=1,2, …,100Sk, k=1,2, …,100Sk, k=1,2, …,100SkSSSkkkkk,, kk==11,,22,, ……,,100100k𕒵k!k𕒵k!k𕒵k!k𕒵k!k𕒵k!k𕒵k!k𕒵k!k𕒵k𕒵kk−𕒵1k!k!kk!!1k1k1k1k1k1k1k111kkk1002100!+100∑k=2|(k2𕒷k+1)Sk|1002100!+100∑k=2|(k2𕒷k+1)Sk|1002100!+100∑k=2|(k2𕒷k+1)Sk|1002100!+100∑k=2|(k2𕒷k+1)Sk|1002100!+100∑k=2|(k2𕒷k+1)Sk|1002100!1002100!1002100210010010022222100!100!100100!!++100∑k=2100∑k=2100∑100∑100∑100100100100100∑∑∑k=2k=2k=2k=2kk==22||||((k2kkk22222−󔼩kk++11))SkSSSkkkkk||||111111222222444444333333A1111111B4444444C3333333D2222222?

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Solution:

Sk=k𕒵k!1𕒵k=1(k𕒵)!, fork>1100∑k=2|(k2𕒷k+1)1(k𕒵)!|