Find the value of <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> n

2nalfq8

2nalfq8

Answered question

2022-07-01

Find the value of n = 1 1 2022 n [ 2021 n 2022 ]

Answer & Explanation

Dalton Lester

Dalton Lester

Beginner2022-07-02Added 12 answers

For the integer part we have that
n = 1 1 2022 n [ 2021 n 2022 ] = 4084440 4088483 0.9990111
and therefore for the fractional part we find (recall that { x } = x [ x ]),
n = 1 1 2022 n { 2021 n 2022 } = n = 1 2021 n 2022 n + 1 n = 1 1 2022 n [ 2021 n 2022 ] = 2021 2022 4084440 4088483 = 4086463 8266912626 0.0004943.
As regards the first sum, we start by using your approach:
n = 1 1 2022 n [ 2021 n 2022 ] = n = 1 1 2022 n [ ( 2022 1 ) n 2022 ] = n = 1 1 2022 n ( k = 0 n 1 ( 1 ) k ( n k ) 2022 n 1 k + [ ( 1 ) n 2022 ] ) = n = 1 1 2022 n ( 2021 n ( 1 ) n 2022 + ( 1 ) n 1 2 ) = 4084440 4088483 .

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