The value of - r -1 n r - r - isA- n -1 -1B- n -1C- n -1 - nD- n -

The correct option is A (n+1)! 1Given sum is S = ∑_r =1^n r× r! Now, nbsp; T_r = r× r! = (r+1 1)× r! ⇒ T_r = (r+1)! r! Now, nbsp; S = ∑_r =1^n r× r!=∑_r =1 ...

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Standard XII

Mathematics

Pascal Triangle

The value of ...

Question

The value of $n \sum r = 1 r \times r!$ is

(n + 1)! - 1

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(n)! - 1

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(n + 1)! - n

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n!

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Solution

The correct option is A $(n + 1)! - 1$
Given sum is $S = n \sum r = 1 r \times r!$
Now,
$T_{r} = r \times r! = (r + 1 - 1) \times r! \Rightarrow T_{r} = (r + 1)! - r!$
Now,
$S = n \sum r = 1 r \times r! = n \sum r = 1 (r + 1)! - r! \Rightarrow S = 2! - 1! + 3! - 2! + \dots + (n + 1)! - n! \Rightarrow S = (n + 1)! - 1$

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The value of n∑r=1r×r! is

The correct option is A (n+1)!−1Given sum is S=n∑r=1r×r! Now, Tr=r×r!=(r+1−1)×r!⇒Tr=(r+1)!−r! Now, S=n∑r=1r×r!=n∑r=1(r+1)!−r!⇒S=2!−1!+3!−2!+⋯+(n+1)!−n!⇒S=(n+1)!−1

The value of $n \sum r = 1 r \times r!$ is