The value of -n-11002 n -1A- 1002B- 1012-100- 992-1D- 1002-100

The correct option is A 100^2∑_n=1^100(2n 1)=∑_n=1^100n^2 (n^2 2n+1) ⇒∑_n=1^100n^2 (n 1)^2 =(1^2 0^2)+(2^2 1^2)+(3^2 2^2)+...+(100^2 99^2) =100^2 Alternate So ...

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Standard XII

Mathematics

Local Maxima

The value of ...

Question

The value of $100 \sum n = 1 (2 n - 1)$ is

100^{2}

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101^{2} - 100

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99^{2} + 1

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100^{2} - 100

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Solution

The correct option is A $100^{2}$
$100 \sum n = 1 (2 n - 1) = 100 \sum n = 1 n^{2} - (n^{2} - 2 n + 1) \Rightarrow 100 \sum n = 1 n^{2} - (n - 1)^{2} = (1^{2} - 0^{2}) + (2^{2} - 1^{2}) + (3^{2} - 2^{2}) + . . . + (100^{2} - 99^{2}) = 100^{2}$

Alternate Solution:
$100 \sum n = 1 (2 n - 1)$
$= 2 100 \sum n = 1 n - 100 \sum n = 1 1$
$= 2 (\frac{100 \times 101}{2}) - 100$
$= 100 \times 101 - 100$
$= 100^{2}$

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Similar questions

The value of $100^{2} - 99^{2}$ is equal to

Q. Find the value of

101^{2} - 99^{2} =

Q. The value of

100^{2} - 99^{2} + 98^{2} - 97^{2} . . . . . . . . . - 3^{2} + 2^{2} - 1^{2}

is:

Q. The least value of

(x + 100)^{2} + (x + 99)^{2} + \dots \dots + (x + 1)^{2} + x^{2} + (x - 1)^{2} + (x - 2)^{2} + \dots . + (x - 100)^{2}

Q. The least value of

{(x + 100)}^{2} + {(x + 99)}^{2} + . . . . + {(x + 1)}^{2} + x^{2} + {(x - 1)}^{2} + {(x - 2)}^{2} + . . . . . . . + {(x - 100)}^{2}

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The value of 100∑n=1(2n−1) is

The correct option is A 1002100∑n=1(2n−1)=100∑n=1n2−(n2−2n+1)⇒100∑n=1n2−(n−1)2=(12−02)+(22−12)+(32−22)+...+(1002−992)=1002 Alternate Solution: 100∑n=1(2n−1) =2100∑n=1n−100∑n=11 =2(100×1012)−100 =100×101−100 =1002

The value of $100 \sum n = 1 (2 n - 1)$ is