How many terms are there in the expansion of (2 + 5x + 3y + 7z)^7 (to the power of 7)?
Answers is 120
Thanks in advance
(2 + 5x + 3y +
7z)^7
= [(2 + 5x) + (3y +
7z)]^7
= (2 + 5x)^7 + A(2 + 5x)^6(3y +
7z) + B(2 + 5x)^5(3y +
7z)^2 + C(2 + 5x)^4(3y +
7z)^3 + D(2 + 5x)^3(3y +
7z)^4 + E(2 + 5x)^2(3y +
7z)^5 + F(2 + 5x)(3y +
7z)^6 + (3y +
7z)7
where A, B, C, D, E, F are constants (can find from calculator, but I'm working mentally online immediately without using one)
Notice that:
All the terms will be in the form x^a * y^b * z^c
where a, b and c varies from 0 to 7, and a + b + c ≤ 7
I'm a bit too stupid to solve it using any igenious way at the moment... so I shall use brute force way first... Hopefully someone can point out a faster route
So,
When a+b+c = 7, we can use the numbers combi of
007, 016, 025, 034, 115, 124, 133, 223
Total number of terms = 3!/2! + 3! + 3! + 3! + 3!/2! + 3! + 3!/2! + 3!/2! = 33 terms
when a+b+c = 6, we can use the numbers combi of
006, 015, 024, 033, 114, 123, 222
Total number of terms = 3!/2! + 3! + 3! + 3!/2! + 3!/2! + 3! + 1 = 31 terms
when a+b+c = 5, we can use the numbers combi of
005, 014, 023, 113, 122
Total number of terms = 3!/2! + 3! + 3! + 3!/2! + 3!/2! = 21 terms
when a+b+c = 4, we can use the numbers combi of
004, 013, 022, 112
Total number of terms = 3!/2! + 3! + 3!/2! + 3!/2! = 15 terms
when a+b+c=3, we can use the numbers combi of
003, 012, 111
Total number of terms = 3!/2! + 3! + 1 = 10 terms
when a+b+c=2, we can use the numbers combi of
002, 011
Total number of terms = 3!/2! + 3!/2! = 6 terms
when a+b+c = 1, only 001. Number of terms = 3
when a+b+c = 0, only 1 term
Total number of terms = 33+31+21+15+10+6+3+1 = 120 terms
Hi,
Based on eagle's expansion, i.e.
(2 + 5x + 3y + 7z)^7
= [(2 + 5x) + (3y + 7z)]^7
= (2 + 5x)^7 + A(2 + 5x)^6(3y + 7z) + B(2 + 5x)^5(3y + 7z)^2 + C(2 + 5x)^4(3y + 7z)^3 + D(2 + 5x)^3(3y + 7z)^4 + E(2 + 5x)^2(3y + 7z)^5 + F(2 + 5x)(3y + 7z)^6 + (3y + 7z)7
We can see that total number of terms = 2 {8 + (7)(2) + (6)(3) + (5)(4)} = 120.
The approach is to appreciate the fact that (a + b)^n has (n + 1) terms
and {(a + b)^n} {(c + d)^m} has (n + 1)(m + 1) terms.
I multiplied by 2 because of symmetry (i.e. last 4 terms in the above expansion is a reflection of the first 4 terms if you observe carefully).
An interesting question, though challenging. Thanks for sharing!
Cheers,
Wen Shih
Thanks for the input.
My friend showed me another answer: 9choose 2 + 8 choose 2 + ... + 2 choose 2 = 120
This question is solved by the use of the multinomial theorem.
However, this theorem is not taught in secondary schools and junior colleges.
Our students are only taught the binomial theorem but not the multinomial theorem
so our students have difficulties to expand it when there are many terms inside the
bracket to a given power.