averyfan1843
Answered

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Simplify
[tex]\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3} +...+\frac{1}{1+2+3+...+99}[/tex]

Sagot :

amerilnoron95

Answer:

65/264 or 0.2462

Step-by-step explanation:

The given series is

(1/1.2.3) + (1/2.3.4) + (1/3.4.5) + ………………

If we denote the series by

u(1) + u(2) + u(3) + u(4) +……………..u(n),

where u(n) is the nth term, then

u(n) = 1/[n(n+1)(n+2)] , n = 1,2,3,4,………n.

which can be written as

u(n) = (1/2) [1/n(n+1) - 1/(n+1)(n+2)] ………………………(1)

In the question, the number of terms n =10, thereby restricting us only to first 10 terms of the series and we have to find the sum for this truncated series. Let S(10) denote the required sum. We have then from (1),

u(1) = (1/2) (1/1.2 - 1/2.3)

u(2) = (1/2) (1/2.3 - 1/3.4)

u(3) = (1/2) (1/3.4 - 1/4.5)

u(4) = (1/2) (1/4.5 - 1/5.6)

u(5) = (1/2) (1/5.6 - 1/6.7)

u(6) = (1/2) (1/6.7 - 1/7.8)

u(7) = (1/2) (1/7.8 - 1/8.9)

u(8) = (1/2) (1/8.9 - 1/9.10)

u(9) = (1/2) (1/9.10 - 1/10.11)

u(10) = (1/2) (1/10.11 - 1/11.12)

Let us now add the terms on LHS and the terms on RHS independently. The sum of LHS is nothing but the sum S(10) of the series up to 10 terms. On the RHS, alternate terms cancel and we are left with only the first and the last term. Therefore,

S(10) = (1/2) (1/1.2 - 1/11.12) = (1/2) (66–1)/132 = [65/(132.2)]

= 65/264

= 0.2462 (correct to four decimal places)

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