sqrt(1) % of sqrt(1) + sqrt(3) % of sqrt(3) + sqrt(5) % of sqrt(5) + .....................+ sqrt(999

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19 Jan 2012 solve

sqrt(1) % of sqrt(1) + sqrt(3) % of sqrt(3) + sqrt(5) % of sqrt(5) + .....................+ sqrt(999999) % of sqrt(999999)

OPtion

1) 25*10^4
2) 25*10^6
3) 25*10^8
4) 64*10^4
5) 64*10^6
6) 64*10^8
7) not possible to solve
8) more than 10^40
9) less than 10^5
10) none of these

Solution

sqrt(x) % of sqrt(x) = x/100
besides it 1+3+5+7+.............+upto n terms = n^2
apply it and get the answer

Correct Option: 3 Best Solution (8)

A.P.Singh   12 years AGO

sqrt(1) % of sqrt(1) + sqrt(3) % of sqrt(3) + sqrt(5) % of sqrt(5) + .....................+ sqrt(999999) % of sqrt(999999)
= 0.01 +0.03 + ......9999.99

which is AP series of n= 500000 terms with
first term a = 0.01
and last term, l = 9999.99

sum of this series = n*(a+l)/2 = 500000* 5000 = 25*10^8

Like? Yes (6) | No (1)  

Ankur mangal   12 years AGO

sqrt(1) % of sqrt (1)=1/10
and
sqrt(3) % of sqrt (3)=3/10
so similarally we can say
(1+3+5........999999)/10
which is an arithmetic progression...

Like? Yes (2) | No   

rahul kumar dhamu   12 years AGO

sqrt(1)% of sqrt(1)=(sqrt(1)/100)*sqrt(1)=1/100
similarly for all we get
(1+3+5+--------+999999)/100
1+3+------999999=(1000000/2)^2=25*(10^10) sum of Arithmetic progression

therefore required sum=(25*(10^10))/100=25*10^8(option 3)

Like? Yes (2) | No   

Shubham   12 years AGO

(1+3+5+7+ ....... + 999999)/100
= 25*10^8

Like? Yes (2) | No   

Shiv Kumar   12 years AGO

sqrt(1)%of sqrt(1)+sqrt(3)%of sqrt(3).....+sqrt(999999)%of sqrt(999999) means (0.01+0.03+0.05.....9999.99) -----1
simplify the equation 1 by multiplying and devide by 100 then it becomes as
(1/100)(1+3+5+........999999) ------ 2
in Equation 2, series is sum of odd numbers between 1 and 1,000,000.
Formula to calculate sum of the "N" (here N is Count)odd numbers is N^2
Between 1 and 10,00,000 count of odd numbers is 5,00,000 (10,00,000/2).
Sum of these 5,00,000 ODD numbers will be 5,00,000^2 = 25*10^10 ------3
By substitute the equation 3 value in equation 2...(1/100)*(25*10^10)
Answer will be 25*10^8

Like? Yes (1) | No (1)  

Rakesh Kumar Kohli   12 years AGO

sqrt(1) % of sqrt(1)= 1/100
sqrt(3) % of sqrt(3)= 3/100
sqrt(5) % of sqrt(5)= 5/100
Thus the problem becomes
(1+3+5+7+..........999999)*1/100
There 500000 terms
Sum = (500000/2)*(1+999999)/100 = 25*10^8

Like? Yes (1) | No (1)  

krishna kumar gupta   12 years AGO

sqrt(n)%sqrt(n)=n/100;
so by this way we can add up a series as follow-
=(1+3+5+......+999999)/100
=500000(1+999999)/(2*100)
=2500*1000000=25*10^8

Like? Yes (1) | No (2)  

Datla Kanthi   12 years AGO

It is AP Series , 1/100(1+3+5+.....+999999)
nth term =a+(n-1)d=999999 so n= 500000
sum of n terms = n/2(a+l)=25*10^10
sum of the given series= 1/100(25*10^10)= 25*10^8

Like? Yes (1) | No (2)