The number of distinct terms in the expansion of (X+Y+Z+W)30
is:
(1)4060 (2)5456 (3)27405 (4)46376
Solution follows here:
This
is application of binomial theorem:
(x+y)n =
nC0 xny0 + nc1 xn-1y +nC2 xn-2y2 +….+
nCn x0yn
Considering
(X+Y) as first term and (Z+W) as second term observe the following binomial
expansions:
(X+Y+Z+W)1
= 1C0 (X+Y)(Z+W)0 + 1C1 (X+Y)0(Z+W)
Number
of terms:
2*1
+ 1*2
(X+Y+Z+W)2
= 2C0 (X+Y)2(Z+W)0 +
2C1 (X+Y)1(Z+W)1 +
2C2 (X+Y)0 (Z+W)2
No.of
terms:
3*1
+
2*2 + 1*3
(X+Y+Z+W)3 =
3C0 (X+Y)3(Z+W)0 + 3C1 (X+Y)2(Z+W) + 3C2
(X+Y)1(Z+W)2 + 3C3 (X+Y)0(Z+W)3
No.of
terms: 4*1 + 3*2
+ 2*3
+ 1*4
If
we observe this pattern, summary is:
(X+Y+Z+W)1 yields
2*1 + 1*2 terms
(X+Y+Z+W)2 yields
3*1 + 2*2 + 1*3 terms
(X+Y+Z+W)3 yields
4*1 + 3*2 + 2*3 +1*4 terms
Going
by this logic,
(X+Y+Z+W)30 yields
31*1
+ 30*2 + .....+ 17*15 + 16*16 +..........+ 1*31 terms
=
2{1*31 + 2*30 +....15*17} + 16*16
=
2{1∑17 n(32-n)} + 16*16
=
2{(1∑17 32*n) - (1∑17 n2)}
+ 256
=
2{32(1∑17 n) - (1∑17 n2)}
+ 256 = 2{32(n(n+1)/2)n=17 - (n(n+1)(2n+1)/6)n=17} +256
=
5456 terms
∴Distinct number of terms in (X+Y+Z+W)30 =
5456
Answer (2)
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