If 1^2 + 2^2 + 3^2 + ................. 100^2 = A & 1^3 + 2^3 + 3^3 + .................. 100^3 = B

M4maths Previous Puzzles

05 Aug 2015 Series

If 1^2 + 2^2 + 3^2 + ................. 100^2 = A
& 1^3 + 2^3 + 3^3 + .................. 100^3 = B
Then
1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101 will be

OPtion

1) A - B
2) A + B
3) A^2 - B^2
4) A^2 + B^2
5) A^2 - B^3
6) A^3 - B^2
7) A + 2B
8) 2A + B
9) A - 2B
10)2A - B

Solution

1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101
= 1^2(1+1) + 2^2(2+1) + 3^2(3+1) + ........... + 100^2(100+1)
= 1^3 + 1^2 + 2^3 + 2^2 + 3^3 + 3^2 + ......... + 100^3 + 100^2
= (1^3 + 2^3 + 3^3)+(1^2 + 2^2 + 3^2 + ....... + 100^2)
= A + B

Option 2)

Correct Option: 2 Best Solution (32)

RAKESH   9 years AGO

1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101

1^2*(1+1) + 2^2*(2+1) + 3^2*(3+1) + 4^2*(4+1) ........ + 100^2*(100+1)

= (1^3 + 2^3 + 3^3 + .................. 100^3) + (1^2 + 2^2 + 3^2 + ................. 100^2) = B+A

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Narendra Gopi   9 years AGO

A=1^2+2^2+3^2+4^2+......100^2
B=1^3+2^3+3^3+4^3+........100^3
A+B=1^2*2+2^2*3+3^2*4+4^2*5+.......100^2*101

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Soma Naresh   9 years AGO

1^2*2+2^2*3+3^2*4+_ _ _ +100^2*101
=1^2*(1+1)+2^2*(2+1)+3^2*(3+1)+_ _ _+100^2*(100+1)
=(1^3+1^2)+(2^3+2^2)+(3^3+3^2)+_ _ _+(100^3+100^2)
=(1^3+2^3+3^3+_ _ _ +100^3)+(1^2+2^2+3^2+_ _ _ +100^2)
=(B+A)

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ANIKET GAURAV   9 years AGO

As 1^2 + 1^3 = 1^2(1+1) = 1^2*2 .. 2^2 + 2^3 = 2^2(1+2) = 2^2*3 ..... andso on upto 100^2 + 100^3 = 100^2 * 101
Hence A+B is right answer

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Apurva Gupta   9 years AGO

a=n^2
b=n^3
c= N^2*(n+1)
=a+b

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Ayush Singhal   9 years AGO

A+B=(1^2+1^3)+(2^2+2^3)+...........(100^2+100^3)
=1^2(1+1)+2^2(1+2)+..........100^2(1+100)
=1^2*2+2^2*3+...............100^2*101
so answer is A+B (2)

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Kundan Gope   9 years AGO

A+B
1^2+1^3=1^2*2
2^2+2^3=2^2*3
3^2+3^3=3^2*4
...
...
100^2+100^3=100^2*101

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SURAJ KUMAR   9 years AGO

Sum of A=100(1+100^2)/2
=500050
Sum of B=100(1+100^3)/2
=50000050
Sum of (A+B)=50500100
Sum of 1^2*2+2^2*3+...+100^2*101=100(2+100^2*101)/2=50500100
Which is equal to A+B

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SOMA NARESH   9 years AGO

1^2*2+2^2*3+3^2*4+_ _ _ +100^2*101
=1^2*(1+1)+2^2*(2+1)+3^2*(3+1)+_ _ _+100^2*(100+1)
=(1^3+1^2)+(2^3+2^2)+(3^3+3^2)+_ _ _+(100^3+100^2)
=(1^3+2^3+3^3+_ _ _ +100^3)+(1^2+2^2+3^2+_ _ _ +100^2)
=(B+A)

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Deepak kumar kumawat   9 years AGO

By adding these terms we will get the answer
Hence A+B is the answer

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Debabrata Dey   9 years AGO

1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101
=1^2(1+1)+2^2(2+1)+3^2(3+1)+4^2(4+1)+........ 100^2(100+1)
=1^3+1^2+2^3+2^2+3^3+2^2+........100^3+100^2
=A+B
so option 2 is correct...

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aditi arora   9 years AGO

taking first three terms of the series:

1^2+2^2+3^2=14=A
1^3+2^3+3^3=36=B
1^2*2+2^2*3+3^2*4=50
A+B=50

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Manuraj Raghuwanshi   9 years AGO

1^2+1^3=1^2(1+1)=1^2*2..

Similarly for any n..

N^2+N^3=N^2(N+1).

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Sriparna Boote   9 years AGO

1^2+2^2+3^2=14
1^3+2^3+3^3=36
14+36=50

and
1^2*2 + 2^2*3 + 3^2*4 =50

So it will be A+B

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shrishti singh   9 years AGO

the given equation gives
Tn=n^2(n+1)
Sn=Summation Tn
=Summation n^3+Summation n^2
=B+A

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SAURABH SINGH   9 years AGO

add the correspomding digits , which are being satisfied by A+B only

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krishnaveni   9 years AGO

because if u verify A+B the result is equal to th e given series result

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Piyush Telang   9 years AGO

Series ( C ) = 2 x 1^2 + 3 x 2^2 + 4 x 3^2 + .................+ 101 x 100^2
- 1^2 + 2^2 + 3^2 + .................+ 100^2
=> C - A = 1 x 1^2 + 2 x 2^2 + 3 x 3^2 +...................+ 100 x 100^2 = B
=> C = A + B

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Nisha Singh   9 years AGO

A is summation of n^2 & B is summation of n^3
the generalised form of the question is the addtion of the summation of A&B


A= summation of N^2 (N=1 to 100)
B=summation of N^3 (N=1 to 100)

A+B = summation N^2(N+1)
=summation (N^3)+summation(N^2)
=A+B

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sai   9 years AGO

we can write B as
1^2*1 + 2^2 *2+ 3^2*3 + ................. 100^2*100
by adding A and B we get the result

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m.lokeswari   9 years AGO

1^2(1+1)+2^2(1+2)+.......+100^2(1+100)
(1^2+2^2+...+100^2)+(1^3+2^3+.......100^3)
A+B

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Rohit Kumar   9 years AGO

1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101
1^2*(1+1) + 2^2*(1+2) + so on-----
(1^2 + 1^3) + (2^2 + 2^3) + so on----
A+B

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Deven Patidar   9 years AGO

1^3 + 2^3 + 3^3 + .................. 100^3 = B
or B= 1^2*1+2^2*2+3^2*3+...........100^2*100
A= 1^2 + 2^2 + 3^2 + ................. 100^2
A+B= 1^2(1+1) + 2^2(1+2) + 3^2(1+3) + ................. 100^2(1+100)
A+B=1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101

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lavanya   9 years AGO

Consider frst two terms
1+4= 5=A
1+8=9=B
A+B= 14= 2+12=14

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Prabhat Burnwal   9 years AGO

1^2*2=1^2(1+1)
similarlily all term can be written in this form

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RAVAL   9 years AGO

1^2+1^3=1^2(1+1)=1^2*2
2^2+2^3=2^2(2+2)=2^2*4
same as .....
A+B

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Subham Mahanta   9 years AGO

here, 1^2 + 2^2 + 3^2 + ................. 100^2 = A
1^3 + 2^3 + 3^3 + .................. 100^3 = B
so, 1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101
= 1^2*(1+1) + 2^2*(2+1) + 3^2*(3+1) + 4^2*(4+1) ........ 99^2*(99+1) + 100^2*(100+1)
=1^2+1^3+2^2+2^3+3^2+3^3+4^2+4^3..........99^2+99^3+100^2+100^3
= (1^2 + 2^2 + 3^2 + ..... 100^2) + (1^3 + 2^3 + 3^3 + ..... 100^3)
= A + B
ans is option (2)

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Sravani   9 years AGO

1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101
=>1^2*(1+1) + 2^2*(2+1) + 3^2*(3+1) + 4^2*(4+1) ........ 99^2*(99+1) + 100^2*(100+1)
=>1^3+1^2+2^3+2^2+.......100^3+100^2
=>(1^2 + 2^2 + 3^2 + ................. 100^2)+(1^3 + 2^3 + 3^3 + .................. 100^3 )
=>A+B

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velmala gangadhar   9 years AGO

1^2 + 2^2 + 3^2 + ................. 100^2 = A
& 1^3 + 2^3 + 3^3 + .................. 100^3 = B
1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*10
can written as..1^2*(1+1)+ 2^2*(2+1) + 3^2*(3+1) + 4^2*(4+1)
then ans..is A+B

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kousik tankala   9 years AGO

3^2*4=(3^2)(3+1)=(3^3)+(1*(3^2)).....separating every term and picking the common terms simplifies it as A+B.

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Prabakaran   9 years AGO

1^2+1^3=1^2*2
2^2+2^3=2^2*3
....
therefore for A+B the given output will be produced

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ashvini kumar   9 years AGO

1^2*2 + 2^2*3 + 3^2*4 + 4^2*5 ........ 99^2*100 + 100^2*101
=E(n^2*(n+1))=E(n^3+n^2)=A+B

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