If an - nn- then -r-1100 ar is equal to

The correct option is C 101! 1 a_n=nn!=((n+1) 1)n!=(n+1)! n! a_1=2! 1! a_2=3! 2! a_3=4! 3! a_100=101! 100! Adding all, we get ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x^2 - y^2)

If an = nn!...

Question

If $a_{n} = n (n!), t h e n \sum_{r = 1}^{100} a_{r}$ is equal to

101!

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100! - 1

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101! - 1

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101! + 1

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Solution

The correct option is C ( $101! - 1$ )
$a_{n} = (n) n! = ((n + 1) - 1) (n!) = (n + 1) n! - n! a_{n} = (n + 1)! - n!$
For $n = 1, 2, 3.....100$ we get
$a_{1} = 2! - 1! a_{2} = 3! - 2! a_{3} = 4! - 3! a_{100} = 101! - 100!$

Adding all, we get
$100 \sum i = 1 a_{n} = 101! - 1$

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Similar questions

Q. If

a = {(100)}^{1 / 100}

and

b = {(101)}^{1 / 101}

then

Q. If

{sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{3 π}{2}

, then the value of

x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}}

Q. If

f (x) = Π_{n = 1}^{100} (x - n)^{n (101 - n)}

, then find

\frac{f (101)}{f^{'} (101)}

$s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2},$ then the value of $x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}} =$

Q. lf

{sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{3 π}{2}

, then

x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}} =

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If an=n(n!),then∑100r=1ar is equal to

The correct option is C (101!−1)an=(n)n!=((n+1)−1)(n!)=(n+1)n!−n!an=(n+1)!−n!For n=1,2,3.....100 we get a1=2!−1!a2=3!−2!a3=4!−3!a100=101!−100!Adding all, we get100∑i=1an=101!−1

If $a_{n} = n (n!), t h e n \sum_{r = 1}^{100} a_{r}$ is equal to

The correct option is C ( $101! - 1$ )
$a_{n} = (n) n! = ((n + 1) - 1) (n!) = (n + 1) n! - n! a_{n} = (n + 1)! - n!$
For $n = 1, 2, 3.....100$ we get
$a_{1} = 2! - 1! a_{2} = 3! - 2! a_{3} = 4! - 3! a_{100} = 101! - 100!$

Adding all, we get
$100 \sum i = 1 a_{n} = 101! - 1$