Number of integral terms in the expansion of (√6 + √10 + √15)^6 is?

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February 13th 2009, 10:31 PM
fardeen_gen

Number of integral terms in the expansion of (√6 + √10 + √15)^6 is?

Number of integral terms in the expansion of (√6 + √10 + √15)^6 is:

A) 3
B) 9
C) 6
D) 10
February 14th 2009, 05:43 AM
Grandad

Integral terms
Hello fardeen_gen

Quote:

Originally Posted by fardeen_gen

Number of integral terms in the expansion of (√6 + √10 + √15)^6 is:

A) 3
B) 9
C) 6
D) 10

In the expansion of $(a+b+c)^6$ , all the terms are in the form $a^p b^q c^r$ , where $p, q, r \ge 0$ and $p+q+r = 6$ .

Now if $a = \sqrt{6}, b = \sqrt {10}$ and $c = \sqrt {15}$ , the only terms that have integer values are when $p, q$ and $r$ are even. (Note that $abc = \sqrt{900} = 30$ , but the only way this combines with other even powers is in the term $a^2b^2c^2$ ; i.e. where $p = q = r = 2$ .)

So:
- if $p=0, q+r = 6$ , and there are 4 possible choices
- if $p = 2, q+r = 4$ , and there are 3 possible choices
- if $p = 4, q+r = 2$ : 2 choices
- if $p=6, q+r = 0$ : 1 choice
Total number of terms, then, is 4+3+2+1 = 10. Answer: D.

Grandad
February 14th 2009, 12:12 PM
Soroban

Hello, fardeen_gen!

A variation of Grandad's explanation . . .

Quote:

Number of integral terms in the expansion of: $\left(\sqrt6 + \sqrt{10} + \sqrt{15}\right)^6$ is:

. . $A)\;3 \qquad B)\;9 \qquad C)\;6 \qquad D)\;10$

In the expansion of $\left(\sqrt{6}+\sqrt{10}+\sqrt{15}\right)^6$ ,

. . the terms are of the form: ${6\choose p,q,r}\left(\sqrt{6}\right)^p\left(\sqrt{10}\right )^q\left(\sqrt{15}\right)^r$ . where $p+q+r \:=\:6$

The term will be an integer if $p,q,r$ are all even.

There are three basic partitions . . .

. . . $\begin{array}{ccc} \{p,q,r\} \:=\:\{0,0,6\} & \text{3 ways} \\ \{p,q,r\} \:=\:\{0,2,4\} & \text{6 ways} \\<br /> \{p,q,r\} \:=\:\{2,2,2\} & \text{1 way} \end{array}$

Therefore, there are 10 integral terms. (answer D)