IIT JEE-AIEEE 2012, Entrance Notifications, Study Tips, Engineering Preparation, Practice Test Papers, Experts Advice

Ask iit jee aieee pet cbse icse state board community

Discussion Response Post to: progression

Forum Index -> Algebra -> View Full Question

Author

Message

Bipin Dubey (13679)

Forum Expert

total posts: 7942
Offline

1 =

           1
⁰^i<k<j^{n         2<=k<=n}^{1<=j<= k-1   0<=i<=j-1}

                  =

        j
   ^2<=k<=n^{1<=j<= k-1}

   =

k(k-1)/2 =

(k²-k)/2
^2<=k<=n^2<=k<=n

=

k²/2 -

k/2
^2<=k<=n^2<=k<=n

=

k²/2 - 1²/2 -

k/2     + 1/2
                   ^1<=k<=n   ^1<=k<=n

   =

k²/2 -

k/2
                     ^1<=k<=n^1<=k<=n

                 = n(n+1)(2n+1)/12 - n(n+1)/4

                 = [n(n+1)/4][(2n+1)/3 - 1]

                 = n(n+1)(2n-2)/6

                 = n(n+1)(n-1)/6

                 = ⁿ⁺¹C₃

Bipin Kumar Dubey Chemical Dept. IIT Kharagpur

Like 0 people liked this

Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info