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Question Number 194135 by cortano12 last updated on 28/Jun/23

determinant (((((23!−23)/(1.1!+2.2!+3.3!+...+21.21!)) =?)))

23!−231.1!+2.2!+3.3!+...+21.21!=?

Answered by Frix last updated on 28/Jun/23

23 [Σ_(j=1) ^n j×j!=(n+1)!−1]

23[∑nj=1j×j!=(n+1)!−1]

Answered by aba last updated on 28/Jun/23

1.1!+2.2!+...+21.21!=(2−1)1!+(3−1)2!+...+(22−1)21! =2!−1!+3!−2!+...+22!−21! =22!−1! A=((23!−23)/(1.1!+2.2!+...+21.21!))=((23(22!−1))/(22!−1!))=23 ✓

1.1!+2.2!+...+21.21!=(2−1)1!+(3−1)2!+...+(22−1)21!=2!−1!+3!−2!+...+22!−21!=22!−1!A=23!−231.1!+2.2!+...+21.21!=23(22!−1)22!−1!=23✓

Answered by MM42 last updated on 28/Jun/23

n×n!=(n+1)!−n! ⇒s=1×1!+2×2!+3×3!+...+21×22!= (2!−1!)+(3!−2!)+(4!−3!)+...+(21!−20!)+(22!−21!)= 22!−1 ⇒((23!−23)/s)=((23(22!−1))/(22!−1))=23 ✓

n×n!=(n+1)!−n!⇒s=1×1!+2×2!+3×3!+...+21×22!=(2!−1!)+(3!−2!)+(4!−3!)+...+(21!−20!)+(22!−21!)=22!−1⇒23!−23s=23(22!−1)22!−1=23✓

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