Question

the value of 50/72+50/90+50/110+......+50/9900 is

Answer 1

72 = 8*9
90 = 9*10
110 = 10*11
and so on
if u take 50 common u are left out with
50(1/8*9 +1/9*10+1/10*11 .........+1/99*100)
=50( 1/8-1/9+1/9-1/10+1/10-1/11.........+1/99-1/100)
=50(1/8-1/100)
=50[(100-8)/800]
=50*92/800
=23/4

Answer 2

Answer:

The value of $\frac{50}{72}+\frac{50}{90}+\frac{50}{110}+...+\frac{50}{9900}$ is $\frac{23}{4}$.

Step-by-step explanation:

Consider the series as follows:

$\frac{50}{72}+\frac{50}{90}+\frac{50}{110}+...+\frac{50}{9900}$

Rewrite the series as follows:

$\frac{50}{8\cdot9}+\frac{50}{9\cdot10}+\frac{50}{10\cdot11}+...+\frac{50}{99\cdot100}$

Take out $50$ common from the numerator of the series.

$50\left[\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{99\cdot100}\right]$

We know that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

Thus,

$50\left[\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{100}\right]$

Observe that inside the bracket all the terms get cancel except the $1^{st}$ and last term.

⇒ $50\left[\frac{1}{8}-\frac{1}{100}\right]$

Simplify as follows:

⇒ $50\left[\frac{100-8}{8\cdot100}\right]$

⇒ $50\left[\frac{92}{800}\right]$

⇒ $\frac{5\cdot92}{80}\right]$

⇒ $\frac{23}{4}$

Therefore, the value of $\frac{50}{72}+\frac{50}{90}+\frac{50}{110}+...+\frac{50}{9900}$ is $\frac{23}{4}$.

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