Below given are two formulae for long summation. 1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2 2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1) The value of (1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be

M4maths Previous Puzzles

05 Feb 2015 Find the value Below given are two formulae for long summation. 1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2 2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1) The value of (1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be: OPtion
1) 2463
2) 2668
3) 2851
4) 2993
5) 3217
6) 3831
7) 3515
8) 3629
9) 3333
10)3967; Correct Option: 9 Best Solution (57)

Supriya Mondal 9 years AGO

From the question we can write,
[(1*2)/100+(2*3)/100+(3*4)/100+.........+(99*100)/100]
=(1/100)*[2+6+12+........+9900]
=(1/100)*[(1+1)+(2+4)+(3+9)+.......+(99+9801)]
=(1/100)*[(1+1^2)+(2+2^2)+(3+3^2)+.......+(99+99^2)]
=(1/100)*[(1+2+3+.......+99)+(1^2+2^2+3^2+........+99^2)]
=(1/100)*[{99*(99+1)/2}+{99*(99+1)*(198+1)/6}]
=3333
So, option 9)3333 is the right answer

Like? Yes (8) | No (3)

Surit Datta 9 years AGO

in this given expression two series is present
{ 2 = 1^2 + 1
6 = 2^2 + 2
12 = 3^2+ 3
20 = 4^2+ 4
......
so on } / 100
so it is the merge up of the above given two series
so by putting value 99
in those two equation and merge up them
we get the value 3333
so option 9

Like? Yes (5) | No (1)

Devendra Marghade 9 years AGO

1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
=(1/100)*2 + (2/100)*3 +(3/100*4)+.......+(99/100)*100
=(1/00)(2+6+12+20+......)
=(1/100)[(1^2+1)+(2^2+2)+(3^2+3)+.........+(99^2+99)]
=(1/100)[(1^2 + 2^2+ 3^2+........+99^2)+(1+2+3+......+99)]
=(1/100)[(1/6)*99*100*199 + (99*100/2)]
=(1/100)[328350+4950]
=(1/100)[333300]
=3333

Like? Yes (4) | No (1)

MRITUNJAY 9 years AGO

subtracting(2 ) from (1)

n(n+1)(2n+1)/6 - n(n+1)/2
after solving equation we get
n(n+1)(n-1)/3
it is general formula for the third series
n=100
100*101*99/3*100=3333
option 9.

Like? Yes (3) | No

VEENUS MAHENDRA 9 years AGO

we can write it as:-
((1*2)+(2*3)+(3*4)........+(99*100))/100

now we separate:-
((1+1^2) + (2+2^2) + (3+3^3) + ........+(n+n^2))/100

now use given formula for summation:-
(n(n+1)/2) + n(n+1)(2n+1)/6)/100

we get:-
n(n+1)(n+2)/300

we put n=99

99*100*101=3333(ans).

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ashvini kumar 9 years AGO

n(n+1)(2n+1)/600+n(n+1)/200=3333,where n=99

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Minal 9 years AGO

1/100[1*2+2*3+...99*100]
1/100[1*(1+1) +2(2+1) +....+99(99+1)]
1/100[(1+2^2+...+99^2) +(1+2+....+99)]
1/100[(99*100*199/6)+99*100/2]

Like? Yes (2) | No

arush gupta 9 years AGO

ans 3333

Like? Yes (1) | No

Deepak Kumar Sahu 9 years AGO

2*(1/100)+3*(2/100)+4*(3/100)+.....+100*(99/100)
=1/100(1*2+2*3+3*4...+99*100)

first solving the bracket portion

1*2+2*3+3*4...+99*100
general formula for each term will be n(n+1)

summation of n(n+1) is the same as summation of n^2 +n.
hence using the summation formula, where n=99,
just plug in 99 to (n/6)(n+1)(2n+1) + (n/2)(n+1) = 333300

then 1/100*(333300)=3333

hence option 9 (ans)

Like? Yes (1) | No

Ajinkya Kulkarni 9 years AGO

=[1*2+2*3+3*4.....]/100

=[2(1+3)+4(3+4).....]/100

=[2*4+4*8+6*12......]/100

=8[1+4+9+16.....]/100 +99*100/100

=8*40425/100 +99

=3234+99

3333

Like? Yes (1) | No

Shanmugapriya.G 9 years AGO

answer for the series is the sum of the above two series divided by 100.
(1%2+2%3+......+99%100)= {(sum of number upto 99) + (sum of square of number upto 99)}/100.
use given formula to find the two series....so answer is 3333

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Swathy Murugan 9 years AGO

option 9

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suman kar 9 years AGO

1% of 2 + 2% of 3 + .....+ 99% of 100 =2/100 + 2*3/100 + 3*4/100 + 4*5/100 +....... +99*100/100 = n*(n+1)/100 where n=1 to 99
=n^2+n/100 where n=1 to 99
=(1^2+2^2+.......+99^2)+(1+2+.......+99)=3333

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NITHYA 9 years AGO

=2(1+3+6+10......+9900)/100
=333300/100
=3333

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apurva katyayni 9 years AGO

option (9)

Like? Yes (1) | No

anusha 9 years AGO

sum of 9 natural nos and sum of squares of 99 nos /100 gives ur ans ie 3333

Like? Yes (1) | No

santosh bagathi 9 years AGO

(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be

= (1*2)/100+ (2*3)/100+(3*4)/100+............+(99*100)/100

= (1/100) *(1*2+2*3+3*4+45*5+............+99*100)

= (1/100)* ([sum for n=1 to 99] n*(n+1))

=(1/100)*[{sum for n=0 to 99} (n^2+n)]

=(1/100)*[(1/6)*n(n+1)(2n+1)]*[ n(n+1)/2] where n=99

substitute and calculate

ans=3333

:)

Like? Yes (1) | No

Anshul Garg 9 years AGO

(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) can be written as:
1*2/100 + 2*3/100 + 3*4/100 +...........99*100/100
=> 1/100(2*(2-1) + 3*(3-1) + ....+100(100-1))

This can be simplified as
1/100[(2^2 + 3^2 ....100^2) - (2+3+....100)]
Also
1/100[(1^2 +2^2 + 3^2 ....100^2) - (1+2+3+....100)]
1/100[ 1/6 n(n+1)(2n+1) - n(n+1)/2]
Where n is 100.
After Solving, we have 3333 as final answer.

Like? Yes (1) | No

Jaspalsingh Parmar 9 years AGO

Option 9 is correct

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Vivek Vishal Anand 9 years AGO

solve
(1/100)(1+1^2 +2+2^2 +3+3^2+............+99+99^2)
which results 3333

Like? Yes (1) | No

Rakshit Shetty 9 years AGO

(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100)
=(1 * 2 + 2 * 3 + 3 * 4 + ........ + 99 * 100)/100
=n(n+1)(n+2)/100
n=99
sum of series = 99*100*101/100 = 3333

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priyadharshini 9 years AGO

ans is 3333

Like? Yes (1) | No

Preethi 9 years AGO

1/100{1*2+2*3+...99*100}
1/100{1*1+2*2+...99*99+1+2+...99}
=3333

Like? Yes (1) | No

krishna 9 years AGO

1%of2+2%of3+--------99%of100

1/100*2+2/100*3+-------------+99/100*100
=1/100(1*2+2*3+3*4+---------------------+97*98+98*99+99*100)
=1/100(2(1+3)+4(3+5)+6(7+9)+-----------+98(97+99)+9900)
=2/100(2^2+4^2+6^2+8^2+--------------+98^2+99*50)
=8/100(1^2+2^2+3^2+4^2+--------------+49^2+99*50/8)
=8/100[(1/6)*(49*50*99)+(99*50)/8]
=(8*99*50)/100[(49/6 + 1/8)]
=3333

or
sigma(n*(n+1))
=sigma (n^2+n)
=sigma n^2+sigma n
=1/6(n(n+1)(2n+1))+[n(n+1)]/2
n=99
then
3333

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ABHINAV PUSHPAM 9 years AGO

Answer is option 9) i.e. 3333

Like? Yes (1) | No

DIVYA SRIVASTAVA 9 years AGO

(sum of n^2 + sum of n)/100

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kumari neha 9 years AGO

We have to find the value of (1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100)
Let S=1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
so S=1*2/100+2*3/100+3*4/100..........+99*100/100
By taking 1/100 common-
S=1/100(1*2+2*3+3*4+.......+99*100)
we can write it as-
S=1/100{∑n*(n+1)} [Limit of ∑ is from n=1 to 99)
so S=1/100{∑n^2+n}
as we know from the given formula
1^2+2^2+.......+99^99=99*100*199/6=328350
and 1+2+3+......+99=99*100/2=4950
so S=1/100(328350+4950)=3333
So answer is 3333

Like? Yes (1) | No

Priyanka Paul 9 years AGO

ans= 1/100 [ {(1/6)*n(n+1)(2n+1)} - {n(n+1)/2} ] , where n= 100

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Satya B 9 years AGO

ans is 3333

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BHASKAR MANDAL 9 years AGO

formula is n(n+1)(n+2)/3 = 99*100*101/3 = 3333
option 9

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Vidhya 9 years AGO

(1*2)/100+........(99*100)/100
Basic formula:
1*2+2*3+...n(n+1)=n(n+1)(n+2)/3
Hence formula for given equ (1*2)/100+........(99*100)/100 is n(n+1)(n+2)/300
(99*100*101)/300=3333

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A M ANKIT KALLURAYA 9 years AGO

by agp formulae

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nishi 9 years AGO

((1*2)/100)+((2*3)/100)+........((99*100)/100)
=1/100(1*2+2*3+3*4......(n-1)n)
=1/100((sigma from 2 to 100)(n^2-n))
=1/100((sum of n^2 from 2 to 100)-(sum of n from 2 to 100)
=1/100(((100*101*201)/6)-1-((100*101)/2)+1)
=3333

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Subha Ganesan 9 years AGO

.08*(1^2+2^3+..................99^2)+99

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souraj sadhukhan 9 years AGO

(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100)
=( (1/100)*2)+((2/100)*3)+.........
=(1/100)[(1^2+1)+(2^2+2)+(3^3+3)+.....]

so sum= n(n+1)/2+n(n+1)(2n+1)/6
=n(n+1)(n+2)/3
=3333 [here n=99]

Like? Yes (1) | No

Sourabh Prasad 9 years AGO

1*2 + 2*3 + 3*4 + 4*5 + .......
2 + 6 + 12 + 20 + ........
(1^2 + 1) + (2^2+2)+ (3^2+3)+ ........
3333

Like? Yes (1) | No

KALPASHREE.S 9 years AGO

taking 100 as common, applying the given 2 rules by expanding the numerals we get= 328350 and 4950
on adding both we get 333300. lastly dividing by 100= 3333

Like? Yes (1) | No

Ankit Pandey 9 years AGO

1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100 = (1*2/100)+(2*3/100)+(3*4/100)+.........................+(99*100/100)

=(1/100)[(1*2)+(2*3)+(3*4)+...........................+(99*100)]

=(1/100)[(2^2-2)+(3^2-3)+(4^2-4)+(5^2-5)+..........+(100^2-100)]

=(1/100)[(1^2 + 2^2 +3^2+........+100^2) - (1+2+3+.........+100)]

=(1/100)[{(100*101*201)/6} - {(100*101)/2}]

=((100*101/200)[(201/3)-1]

=101*198/3

=101*33
=3333

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Nupur 9 years AGO

1/100( 2*1 + 2*3 + 3*4 + 4*5 + 5*6..... 99 *100)
= 1/100( 2(4) + 4(8) + 6(12) + 8(16).....) + 99
= 1/100{ 2( 2^2 + 4^2 + 6^2 ....... 98^2)} + 99
= 1/100*2*4( 1 + 2^2 + 3^2 ....+ 49^2) + 99
= 3234 + 99
= 3333
HENCE THE SOLUTION IS 3333

Like? Yes (1) | No

jyotsna kumari 9 years AGO

2*1/100 + 2*3/100 + 3*4/100 +......+99*100/100
1/100[2+6+12+20+........+9900]
1/100[(1^2+1)+(2^2+2)+(3^2+3)...+(99^2 + 99)]
1/100[{1^2 + 2^2 +...+99^2}+{1+2+3+...+99}]
1/100[(99*199*100/6)+(99*100/2)]
99*404/12 =3333

Like? Yes (1) | No

Meenakshi 9 years AGO

=>1% of 2 + 2% of 3+ 3% of 4+......+99% of 100
=>2/100+6/100+12/100+..........+9900/100
=>1/100(2+6+12+20+30+......+9900)
=>1/100((1^2+1)+(2^2+2)+(3^2+3)+........(99^2+99))
=>1/100((1^2+2^2+3^2+...+99^2)+(1+2+3+....+n))
=>1/100((99*100*199/6)+(99*100/2))
=>1/100(328350+4950)
=>1/100(333300)
=>3333
So answer is option 9)

Like? Yes (1) | No

Avani Patoria 9 years AGO

Here we can take ;
1% of 2 +2% of 3+……+99% of 100 =( ∑(n^2+n))/100
=(∑n^2+∑n)/100
Take n=99 and By using given 2 formulas we can get the answer = 3333

Like? Yes (1) | No

sumit kumar 9 years AGO

substrate formula 1 from formula 2
we get 2+6+12+......+n^2-n=n(n+1)/2[1/3(2n+1)-1] .....(1)
from given question taking 1/100 common
we get 1/100(2+6+12+20+.....+10^2-10)............(2)
from equation (1) and (2)
n=10 then answer is from equation (1) is 3333

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Jitendra Kumar 9 years AGO

(1.2+2.3+...........+99.100)/100

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b.subbarao 9 years AGO

1/100*2+2/100*3+.......................99/100*100
1/00(1*2+2*3+3*4+.............99*100)
1/100(99*100*199/6+99*100/2)
1/100(328350+4950)=1/100(333300)=3333
option 9 is coreect

Like? Yes (1) | No

JEETENDER KALACHAND KEWLANI 9 years AGO

1+2+3+...+n = n(n+1)/2 ...1
1^2 + 2 ^2 +....+n^2=n(n+1)(2n+1)/6 ...2
subtract equation 1 from 2 , we get
(1^2-1) + (2^2-2)+(3^2-3)+....+(n^2-n)=[n(n+1)(2n+1)-3n(n+1)]/6
1(1-1) + 2(2-1) + 3(3-1)+.....+n(n-1)=n(n+1)[2n+1-3]/6 ...3

now L.H.S of equation 3 is similar to the question sequence :
[0*1 + 1*2 + 2*3....+ (n-1)*n]/100 =

substituting value of eq 3 in above eq.
[n(n+1)(2n-2)/6]/100...
put n=100
and we get 3333

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priyanka dasari 9 years AGO

1percent of 2 is written as(addition of1&1^2)/100, 2percent of 3 is written as (addition of 2&2^2)/100........... 99percent of 100 is written as(addition of99&99^2)/100.
i.e. sum of first 99 natural numbers plus sum of squares of first 99 natural numbers divided with 100 will give the answer.

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shreepati rout 9 years AGO

1/100{(1*2)+(2*3)+(3*4+)..........+(99*100)}
=1/100[{1*(1+1)}+{2*(2+1)}+{3*(3+1)}..........{99*(99+1)}]
=1/100[{1^2+2^2+3^2........99^2}+{1+2+3+4........99}]
=3333

Like? Yes (1) | No

Paras Jain 9 years AGO

its 3333 Ans.

Like? Yes (1) | No

Abhishek singh 9 years AGO

1/100(1*2+2*3+3*4.........)
=>1/100(2+6+12+20+.....)
=>1/100(1^2+1+2^2+2+3^2+3.....)
using formula
1/100(333300)=3333

Like? Yes (1) | No

prakash kumar 9 years AGO

1/100((1^1 + 2^2 + 3^3 + ..........+100^100) - (1+2+3+.......+100))
= (101*201*100)/600 - (101*201)/200
= 101/2(201/3 - 1) = 101*33 = 3333

Like? Yes (1) | No

Sanu Panji 9 years AGO

N(N+1)/100
=(N^2+N)/100
=[(1+2+3+.....+N)+(1^2+2^2+3^2+.....+N^2)]/100
=[N(N+1)+1/6N(N+1)(2N+1)]/100
=3333 [as here N=99]

Like? Yes (1) | No

Saurabh Kumar Gupta 9 years AGO

The given pattern can be written as:
(1/100)*2 + (2/100)*3 + (3/100)*4 + ..... + (99/100)*100
It is of the form n(n+1)/100
i.e. (n^2 + n)/100 where n = (1 - 99)
So using the formula, the summation becomes
= (328350 + 4950)/100
= 333300/100
= 3333

Like? Yes (1) | No (1)

Gurram Mahendra ranganath 9 years AGO

1/100*(99*100*101/3)=3333

Like? Yes (1) | No (1)

sravani 9 years AGO

It can be written as 1/100(1*2+2*3+3*4...+99*100)=1/100(n*(n+1)*(n+3)/3)

so,99*100*101 / 100*3 =3333

Like? Yes (1) | No (1)

anuj gupta 9 years AGO

3333

Like? Yes (1) | No (1)

RAKESH 9 years AGO

1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100

= ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100

= (1/100)* sigma(n^2+n) [n=1 to n=99]

= 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ]
put n=99
= 3333

Like? Yes (1) | No (2)

Dimple

India

Punjab

Time: 00:01:33

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Below given are two formulae for long summation. 1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/

M4maths Previous Puzzles

05 Feb 2015 Find the value

Below given are two formulae for long summation.
1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2
2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1)
The value of
(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be

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