@mihir66 said:1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?A) 100.2^100B) 99.2^100 + 1C) 99.2^99 + 99D) 2^100
b ?
general term = n*2^(n - 1)
for n = 1....general term = 1
now check with options......b satisfies as general term = (n - 1)*2^(n) + 1
@ChirpiBird said:blue colored part. wasn't that obvious? the parts are shown to make calculations easy. pi/4*(1/2)^2 + 1/4 - pi/4*(1/2)^2 = 1/4.
where does 1/4 come from?
dont mind explaining fully
@mihir66 said:1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?A) 100.2^100B) 99.2^100 + 1C) 99.2^99 + 99D) 2^100
B
@iLoveTorres said:
@mihir66 said:1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?
A) 100.2^100
B) 99.2^100 + 1
C) 99.2^99 + 99
D) 2^100
arithmatico geometric series
s=1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99
2s=(-1-2^1........-2^99)+100.2^100
solving we get
99*2^100+1
Questions
@DeAdLy said:where does 1/4 come from?dont mind explaining fully
yes sure.
see figure. i have added area numbers.
area 1 + area 2 will give u the required answer.
area 1 is basically quarter circle with radiius 1/2
(pi*(1/2)^2)/4
area 2 = area of the quarter part of square - quarter circle
=1/4 - (pi*(1/2)^2)/4
see figure. i have added area numbers.
area 1 + area 2 will give u the required answer.
area 1 is basically quarter circle with radiius 1/2
(pi*(1/2)^2)/4
area 2 = area of the quarter part of square - quarter circle
=1/4 - (pi*(1/2)^2)/4
Questions
@abhishek.2011 said:arithmatico geometric seriess=1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 2s=(-1-2^1........-2^99)+100.2^100solving we get99*2^100+1
Please explain how did u derived "2s=(-1-2^1........-2^99)+100.2^100"
@ChirpiBird said:yes sure.see figure. i have added area numbers.area 1 + area 2 will give u the required answer.area 1 is basically quarter circle with radiius 1/2(pi*(1/2)^2)/4area 2 = area of the quarter part of square - quarter circle=1/4 - (pi*(1/2)^2)/4
p,q,r are non negative integers such that p + q + r = 10 the maximum value of pqr + pr + qr + pq is...
for the above question, anyone please give a proper approach
@mihir66
s = 1 + 2*2^1 + 2*2^2 + ..............+ 100*2^99...i
now multiply this eqn by 2
2s = 1*2 + 2*2^2 + 2*2^3 + ..........+ 99*2^99 + 100*2^100...ii
i - ii
s - 2s = 1 + 2^1 + 2^2 +..........+ 2^99 - 100*2^100
-s = 1( 2^100 - 1)/(2 - 1) - 100*2^100
s = 100*2^100 - (2^100 - 1)
s = 100*2^100 - 2^100 + 1
s = 2^100*(100 - 1) + 1
s = 99*2^100 + 1
Lagaan is levied on the 60% of the cultivated land. The revenue dept collected total Rs. 384000 through the lagaan from the village of Sukhiya. Sukhiya, a very rich farmer, paid only Rs. 480 as lagaan. What is the percentage of total land of Sukhiya over the total taxable land of the village?
@mihir66 :
s=1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99
2s= 1.2 + 2.2^2 + 3.2^3 + ................ + 99.2^99 + 100.2^100
Subtracting we get
-s=1+ 2^1 + 2^2 + 2^3 + ...... + 2^99 - 100.2^100
s=100.2^100 - (1 + 2^1 + 2^2 + 2^3 + ... +2^99)
s=100.2^100 - (2^100 - 1)
s=99.2^100 + 1
In a class of 100 students 70 passed in physics, 62 passed in mathematics, 84 passed in english and 82 passed in chemistry. 37 students passed in all 4 subjects. How many maximum students could have failed all four subjects?
(a) 12 (b) 17 (c) can not be determined (d) none of the foregoing
(a) 12 (b) 17 (c) can not be determined (d) none of the foregoing
@Shrutim90 said:Lagaan is levied on the 60% of the cultivated land. The revenue dept collected total Rs. 384000 through the lagaan from the village of Sukhiya. Sukhiya, a very rich farmer, paid only Rs. 480 as lagaan. What is the percentage of total land of Sukhiya over the total taxable land of the village?
12.50 % ?
@jain4444 said:In a class of 100 students 70 passed in physics, 62 passed in mathematics, 84 passed in english and 82 passed in chemistry. 37 students passed in all 4 subjects. How many maximum students could have failed all four subjects?(a) 12 (b) 17 (c) can not be determined (d) none of the foregoing
Option d