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 x^n-1y +nC2 x^n-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)¹c^o + 1C1 (a+b)⁰c¹       

Number of terms:              2*1        +      1*1  

(a+b+c)²     = 2C0  (a+b)² c^o + 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 c^o + 3C1  (a+b)²c + 3C2  (a+b)¹c² + 3C3  (a+b)⁰c³

 = 3C0  (a³+b³+3a²b+3ab²) c^o + 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

Answer (1)