According to the empirical rule, if the data form a "bell-shaped" normal distribution, approximately _______ percent of the obse
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It looks like you have to find the value of the sum,
so that the <em>n</em>-th term in the sum is
for 1 ≤ <em>n</em> ≤ 180.
We can then write the sum as
Break up the summand into partial fractions:
Combine the fractions into one with a common denominator and set the numerators equal to one another:
Expand the right side and collect terms with the same power of <em>n</em> :
Then
<em>a</em> + <em>b</em> + <em>c</em> = 0
43<em>a</em> + 42<em>b</em> + 41<em>c</em> = 0
462<em>a</em> + 440<em>b</em> + 420<em>c</em> = 1
==> <em>a</em> = 1/2, <em>b</em> = -1, <em>c</em> = 1/2
Now our sum is
which is a telescoping sum. If we write out the first and last few terms, we have
<em>C</em> = 1/(2×21) - 1/22 <u>+ 1/(2×23)</u>
… … + 1/(2×22) - 1/23 <u>+ 1/(2×24)</u>
… … <u>+ 1/(2×23)</u> - 1/24 <u>+ 1/(2×25)</u>
… … <u>+ 1/(2×24)</u> - 1/25 <u>+ 1/(2×26)</u>
… … + … - … + …
… … <u>+ 1/(2×198)</u> - 1/199 <u>+ 1/(2×200)</u>
… … <u>+ 1/(2×199)</u> - 1/200 + 1/(2×201)
… … <u>+ 1/(2×200)</u> - 1/201 + 1/(2×202)
Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,
<em>C</em> = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)
which we simplify further to
<em>C</em> = 1/42 - 1/44 - 1/402 + 1/404
<em>C</em> = 1,115/1,042,118 ≈ 0.00106994