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# help

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Find the first of 100 consecutive odd integers whose sum is 100^{100}.

Dec 12, 2019

### 5+0 Answers

#1
+88
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Let the first integer be x so we don't have to answer the question after we have finished the problem. So the 100th term will be x+99. So the sum of these 100 integers are 100x + 4950. Since the sum are 100^100. So 100x is 100^100 - 4950 so x will be 10^99 - 99/2.

~~GF

Dec 12, 2019
#2
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∑[(1E198 - (2n - 1)), n, 1, 199] =1E200 =100^100

Dec 12, 2019
#3
+108626
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Gore, I know you are very capable but on this occasion you did not read the question properly.

$$S_n=\frac{n}{2}(2a+(n-1)d)\\~\\ 10^{10}=\frac{100}{2}(2a+(100-1)*2)\\~\\ 10^{10}=\frac{100}{2}(2a+99*2)\\~\\ 10^{10}=100(a+99)\\~\\ 10^{8}=a+99\\~\\ a=10^{8}-99\\~\\$$

Check

$$S_n=\frac{n}{2}(2a+(n-1)d)\\~\\ S_{100}=\frac{100}{2}(2(10^8-99)+99*2)\\~\\ S_{100}=100(10^8-99+99)\\~\\ S_{100}=100(10^8)\\~\\ S_{100}=100^{10}$$

Of course this is not a great check because I have used exactly the same formula as I did when i worked it out.

S_n=\frac{n}{2}(2a+(n-1)d)\\~\\

10^{10}=\frac{100}{2}(2a+(100-1)*2)\\~\\

10^{10}=\frac{100}{2}(2a+99*2)\\~\\

10^{10}=100(a+99)\\~\\

10^{8}=a+99\\~\\

a=10^{8}-99\\~\\

Dec 12, 2019
#4
+1

Hi Melody: I think you made a typo in the SUM of the sequence. It is 100^100 not 10^10.

100/2 * [2F + 2*(100 - 1)]=100^100, solve for F

F =195 x 9 + 901 = 198 digits long =~ 1 x 10^198

Dec 12, 2019
edited by Guest  Dec 12, 2019
#5
+108626
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You are right, So I did.

Thanks for picking that up.  :)

Melody  Dec 12, 2019