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

• Feb 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
• Feb 14th 2009, 05:43 AM
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.

• Feb 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 . . .

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

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