Question Number 134605 by Khalmohmmad last updated on 05/Mar/21 | ||
Answered by Ñï= last updated on 05/Mar/21 | ||
$${f}\left(\mathrm{2}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}} {\sum}}{k}!=\mathrm{1}!+\mathrm{2}!=\mathrm{3} \\ $$$${g}\left({f}\left(\mathrm{2}\right)\right)={g}\left(\mathrm{3}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\left({k}+\mathrm{1}\right)=\mathrm{2}+\mathrm{3}+\mathrm{4}=\mathrm{9} \\ $$ | ||