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3 years ago

Let x be the sum of the following 2000 numbers

Mathematics

2 answers:

aev [14]3 years ago

8 0

Answer:

See attached image for the sum computed using wolfram alpha, raw value is: 49382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382715160

goblinko [34]3 years ago

4 0

Answer:

Step-by-step explanation:

First, i divide all terms by 4

sequence is 1,11,111,1111,....,1111...1111 (with 2000 digits)

$u_1=1\\\\u_2=11=10+1=\dfrac{10^2-1}{9}\\\\u_3=111=100+10+1=\dfrac{10^3-1}{9}\\\\u_4=1111=1000+100+10+1=\dfrac{10^4-1}{9}\\....\\u_n=\dfrac{10^n-1}{9}\\\\\\\displaystyle \sum_{i=1}^n\ u_1 =\dfrac{10^1-1}{9}+\dfrac{10^2-1}{9}+\dfrac{10^3-1}{9}+...+\dfrac{10^n-1}{9}\\=\dfrac{1}{9} *(10+10^2+10^3+....+10^n+1-1-(1+1+1+1...+1)\ )\\\\=\dfrac{1}{9}*( \frac{10^{n+1}-1}{9} -(n+1)\ )$

$\boxed{4+44+444+....+444...4=\dfrac{4}{9}*( \dfrac{10^{2001}-1}{9} -2001\ )}\\\\\\\approx{4,9382716049382716049382716049383*10^{1999} }$

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Let x be the sum of the following 2000 numbers​

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