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Question Number 134605 by Khalmohmmad last updated on 05/Mar/21

Answered by Ñï= last updated on 05/Mar/21

f(2)=Σ_(k=1) ^2 k!=1!+2!=3  g(f(2))=g(3)=Σ_(k=1) ^3 (k+1)=2+3+4=9

$${f}\left(\mathrm{2}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}} {\sum}}{k}!=\mathrm{1}!+\mathrm{2}!=\mathrm{3} \\ $$$${g}\left({f}\left(\mathrm{2}\right)\right)={g}\left(\mathrm{3}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\left({k}+\mathrm{1}\right)=\mathrm{2}+\mathrm{3}+\mathrm{4}=\mathrm{9} \\ $$

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