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Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:

What I tried to do was make pairs of two terms and the expand it as a binomial expression and then again expand the binomial in the resulting series which gave me an expression with lot of unknowns and I got stuck.

Any help would be appreciated. Thanks.

P Vanchinathan
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Gokul
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3 Answers3

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Hint: What happens if you evaluate your expression in $x=1,y=0,z=1,w=1$? And in $x=0,y=0,z=1,w=1$?

Jack D'Aurizio
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  • I understand that if I put $y=0, x=1, z=1, w=1$, I will get the sum of coefficients which will not have any terms with $y$ in it and if I put both $x=y=0$, I will get the sum of coefficient of terms not containing both $x$ and $y$. Pardon my ignorance, but I don't understand how that'll help me. – Gokul Apr 08 '15 at 12:16
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    @Gokul Hint: What do you get when you subtract those? – Kyle Apr 08 '15 at 15:12
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$$(x+y+z+w)^6=\{y+(x+w+z)\}^6=\cdots+(x+w+z)^6$$

$$(x+w+z)^6=\sum_{r=0}^6\binom6rx^{6-r}(w+z)^r$$

We need $r\ne6$

The sum of the reuqired coefficients should be $\sum_{r=0}^5\binom6r(1+1)^r$ (setting $x=w=z=1$)

$$=\sum_{r=0}^6\binom6r2^r-\binom662^6=(1+2)^6-2^6$$

lab bhattacharjee
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4

Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:

Sum of terms with no y : $3^6$ (y=0 rest all 1)
Sum of terms with no y and no x: $2^6$ (x,y=0 rest all 1)
Sum of terms with no y but x: $3^6-2^6=665$ (subtract the above)

RE60K
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