How do you find the partial sum of #Sigma (2n-1)# from n=1 to 400?

Answer 1

#sum_(n=1)^400 (2n-1) = 16000#

We have that using the distributive property of the sum:

#sum_(n=1)^N (2n-1) = 2 sum_(n=1)^N n -sum_(n=1)^N 1#
The sum of the first #N# integers is given by Gauss' formula:
#sum_(n=1)^N n = (N(N+1))/2#
while the sum of #N# times the unity is clearly #N#:
#sum_(n=1)^N 1 = N#

So:

#sum_(n=1)^N (2n-1) = 2(N(N+1))/2 -N = N^2+N-N = N^2#
For #N=400#:
#sum_(n=1)^400 (2n-1) =400^2 = 16000#
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Answer 2

To find the partial sum of the series ( \sum_{n=1}^{400} (2n - 1) ), you can use the formula for the sum of an arithmetic series:

[ S_n = \frac{n}{2}(a_1 + a_n) ]

where: ( S_n ) is the sum of the first ( n ) terms, ( a_1 ) is the first term of the series, ( a_n ) is the nth term of the series.

In this series, ( a_1 = 2(1) - 1 = 1 ) and ( a_{400} = 2(400) - 1 = 799 ).

Substituting these values into the formula:

[ S_{400} = \frac{400}{2}(1 + 799) = \frac{400}{2} \times 800 = 200 \times 800 = 160000 ]

So, the partial sum of the series ( \sum_{n=1}^{400} (2n - 1) ) is 160,000.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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