Progressions

[2010gmat]

1. Find the value of (100*1) + ( 99*2) + (98*3) + ...... + (2*99) + (1*100)

a.) 12850
b.) 272700
c.) 23250
d.) 50500
e.) none of these

2.) Find the sum of first 20 terms of the series 1.2^2 + 2.3^2 + 3.4^2+ .....
a.) 47980
b.) 46180
c.) 47160
d.) 48180
e.) None of these

[uttam.albela]

1. Find the value of (100*1) + ( 99*2) + (98*3) + ...... + (2*99) + (1*100)

a.) 12850
b.) 272700
c.) 23250
d.) 50500
e.) none of these

2.) Find the sum of first 20 terms of the series 1.2^2 + 2.3^2 + 3.4^2+ .....
a.) 47980
b.) 46180
c.) 47160
d.) 48180
e.) None of these

Q1. General term is (101-n)*n = 101n - n*n
Sum of n consecutive integers starting with 1= n*(n+1)/2
Sum of squares of n consecutive integers starting with 1 = n(n+1)(2n+1)/6
Use these 2 formulas to get the answer.
Q2. General term is n*(n+1)*(n+1)=n3+2n2+n
Sum of cubes of n consecutive integers starting with 1= square of [n*(n+1)/2]
Use this formula along with above two formulas to solve it.

[2010gmat]

awesome dude!!!

so the trick is to simply find the general term

Thanks a lot