If 1-r-022rr-2-1- r-k- - then the value of k is

1+∑_r=0^22{r(r+2)+1}· r! =1+∑_r=0^22{(r+1 1)(r+2)+1}r! =1+∑_r=0^22[(r+2)! (r+1)!] =1+ (2! 1!)+(3! 2!)+⋯+(24! 23!) nbsp;= nbsp;1+24! 1!=24! Therefore, ...

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Standard XII

Mathematics

Higher Order Derivatives

If 1+∑r=022rr...

Question

If $1 + 22 \sum r = 0 {r (r + 2) + 1} \cdot r! = k!,$ then the value of $k$ is

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Solution

The correct option is C $24$
$1 + 22 \sum r = 0 {r (r + 2) + 1} \cdot r! = 1 + 22 \sum r = 0 {(r + 1 - 1) (r + 2) + 1} r! = 1 + 22 \sum r = 0 [(r + 2)! - (r + 1)!] = 1 + (2! - 1!) + (3! - 2!) + \dots + (24! - 23!) = 1 + 24! - 1! = 24!$
Therefore, $k = 24$

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If 1+22∑r=0{r(r+2)+1}⋅r!=k!, then the value of k is

The correct option is C 241+22∑r=0{r(r+2)+1}⋅r!=1+22∑r=0{(r+1−1)(r+2)+1}r!=1+22∑r=0[(r+2)!−(r+1)!]=1+(2!−1!)+(3!−2!)+⋯+(24!−23!)=1+24!−1!=24! Therefore, k=24

If $1 + 22 \sum r = 0 {r (r + 2) + 1} \cdot r! = k!,$ then the value of $k$ is

The correct option is C $24$
$1 + 22 \sum r = 0 {r (r + 2) + 1} \cdot r! = 1 + 22 \sum r = 0 {(r + 1 - 1) (r + 2) + 1} r! = 1 + 22 \sum r = 0 [(r + 2)! - (r + 1)!] = 1 + (2! - 1!) + (3! - 2!) + \dots + (24! - 23!) = 1 + 24! - 1! = 24!$
Therefore, $k = 24$