Question

1/1*3*5 +1/1*4 +1/3*5*7 +1/4*7 +1/5*7*9 +1/7*10 + ------------------upto 20terms

Answer 1

Answer:

          6070 / 14973  ≈  0.4054

Step-by-step explanation:

The given sequence of terms is actually made up by taking terms alternately from two different sequences.  So taking the sum to 20 terms is just taking 10 terms from one sequence and 10 terms from the other:

$\displaystyle\frac1{(1)(3)(5)}+\frac1{(1)(4)}+\dots\text{to 20 terms}\\\\=\frac1{(1)(3)(5)}+\frac1{(3)(5)(7)}+\frac1{(5)(7)(9)}+\dots+\frac1{(19)(21)(23)}+\\\mbox{}\ \ \ \ \ \frac1{(1)(4)}+\frac1{(4)(7)}+\frac1{(7)(10)}+\dots\frac1{(28)(31)}$

Now deal with each of the two parts separately.

The first ten terms have the form:

$\displaystyle\frac1{(n-2)n(n+2)} = \frac18\left(\frac1{n-2}-\frac2n+\frac1{n+2}\right)$

So the sum is

$\displaystyle\frac1{(1)(3)(5)}+\frac1{(3)(5)(7)}+\frac1{(5)(7)(9)}+\dots+\frac1{(19)(21)(23)}\\\\=\frac18\left(\frac11-\frac23+\frac15+\frac13-\frac25+\frac17+\frac15-\frac27+\frac19+\dots+\frac1{19}-\frac2{21}+\frac1{23}\right)\\\\=\frac18\left(\frac11-\frac13-\frac1{21}+\frac1{23}\right)\\\\=\frac{(21)(23)-(7)(23)-23+21}{(8)(21)(23)}\\\\=\frac{40}{483}$

The last ten terms have the form:

$\displaystyle\frac1{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right)$

So the sum is

$\displaystyle\frac1{(1)(4)}+\frac1{(4)(7)}+\frac1{(7)(10)}+\dots\frac1{(28)(31)}\\\\=\frac13\left(\frac11-\frac14+\frac14-\frac17+\frac17-\frac1{10}+\dots+\frac1{28}-\frac1{31}\right)\\\\=\frac13\left(\frac11-\frac1{31}\right)\\\\=\frac{10}{31}$

Adding these together, the sum of the first 20 terms in the original sequence is then

$\displaystyle\frac{40}{483}+\frac{10}{31}=\frac{6070}{14973}$

As a decimal, this is approximately 0.4054.

Hope that helps.

Answer 2

Answer:

Step-by-step explanation:

The given sequence of terms is actually made up by taking terms alternately from two different sequences.  So taking the sum to 20 terms is just taking 10 terms from one sequence and 10 terms from the other:

Now deal with each of the two parts separately.

The first ten terms have the form:

So the sum is

The last ten terms have the form:

So the sum is

Adding these together, the sum of the first 20 terms in the original sequence is then

As a decimal, this is approximately 0.4054.