## Monday, 24 October 2011

### Algebra-7 (CAT-2008)

What is the number of distinct terms in the expansion of (a+b+c)20?
(1)231              (2)253    (3)242   (4)210          (5) 228
Solution follows here:

Solution:
This is application of binomial theorem:
(x+y)n = nC0 xn y0 + nc1 xn-1y +nC2 xn-2y2 +….+ nCn yn
Considering (a+b) as first term and ‘c’ as second term observe the following binomial expansions:

(a+b+c)1                    = 1C0 (a+b)1co + 1C1 (a+b)0c1
Number of terms:              2*1        +      1*1

(a+b+c)2     = 2C0  (a+b)2 co + 2C1  (a+b)1c1 + 2C2  (a+b)0c2
= 2C0 (a2+2ab+b2)*1 + 2C1 (a+b)*c     + 2C2 (1*c2
No. of terms:                3*1            +           2*1         +       1*1

(a+b+c)3 = 3C0  (a+b)3 co + 3C1  (a+b)2c + 3C2  (a+b)1c2 + 3C3  (a+b)0c3
= 3C0  (a3+b3+3a2b+3ab2) co + 3C1  (a2+b2+2ab)c + 3C2  (a+b)c2 + 3C3 1* c3
No.of terms:     4*1       +            3*1                 +           2*1      +     1*1

If we observe this pattern, summary is:
(a+b+c)1 yields        2*1 + 1*1                             = 2+1                         terms
(a+b+c)2 yields        3*1 + 2*1 + 1*1                = 3+2+1        terms
(a+b+c)3 yields        4*1 + 3*1 + 2*1 + 1*1     = 4+3+2+1   terms
Going by this logic,
(a+b+c)n yields (n+1)+n+(n-1)+……..+3+2+1 = ∑(n+1)        terms
= (n+2)(n+1)/2                          (as we know that Σn = (n+1)n/2)
Distinct number of terms in (a+b+c)20 = (20+2)(20+1)/2 = 11*21 = 231