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01 Sep 2020, 08:00
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B
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Difficulty:
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Question Stats:
41% (01:52) correct
59%
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wrong
based on 399
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What is the coefficient of \(m^9\) if you expand the following product:
\((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111})\)
A. 7
B. 8
C. 9
D. 10
E. 11
M37-87
\((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111})\)
A. 7
B. 8
C. 9
D. 10
E. 11
|
M37-87
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02 Sep 2020, 02:17
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Thought process:
First thing that should come to your mind when you look at such seemingly heavy-duty questions is – what concept is it testing?
Powers and Exponents!
So, what do I know about “Powers and Exponents” that I can possibly use here? Well, I can immediately think of some properties of exponents.
Property (powers add when same bases are multiplied): a^x * a^y = a^x+y
But, why did I think of this property? Why not other exponent properties? Because, I am given a product or multiplication of some terms and asked to find coefficient of m^9.
Cool, now that we have a starting point, let’s see what we can do with this property. Fundamental question that we need to answer is “How can we form a power of 9?”
Here are some of the ways we can get a SUM of 9 in power:
(1) 0 + 9 = 9 ---> 1, m^9 when multiplied, will give = m^9
(2) 1 + 8 = 9 ---> m, m^8 when multiplied, will give = m^9
(3) 2 + 7 = 9 ---> m^2, m^7 when multiplied, will give = m^9
(4) 3 + 6 = 9 ---> m^3, m^6 when multiplied, will give = m^9
(5) 4 + 5 = 9 ---> m^4, m^5 when multiplied, will give = m^9
(6) 1 + 2 + 6 = 9 ---> m, m^2, m^6 when multiplied, will give = m^9
(7) 1 + 3 + 5 = 9 ---> m, m^3, m^5 when multiplied, will give = m^9
(8) 2 + 3 + 4 = 9 ---> m^2, m^3, m^4 when multiplied, will give = m^9
These are the only combination of products of power of m which would lead to m^9. Note that only first nine terms from (1+m) to (1+m^9) would give us m^9 terms. All remaining terms will give higher powers of m and hence not of any use for us.
Since each of the m^9 term has a coefficient = 1, so adding all eight m^9 terms, we get = 8*m^9
So, coefficient of m^9 = 8
Option B (correct)
Posted from my mobile device
First thing that should come to your mind when you look at such seemingly heavy-duty questions is – what concept is it testing?
Powers and Exponents!
So, what do I know about “Powers and Exponents” that I can possibly use here? Well, I can immediately think of some properties of exponents.
Property (powers add when same bases are multiplied): a^x * a^y = a^x+y
But, why did I think of this property? Why not other exponent properties? Because, I am given a product or multiplication of some terms and asked to find coefficient of m^9.
Cool, now that we have a starting point, let’s see what we can do with this property. Fundamental question that we need to answer is “How can we form a power of 9?”
Here are some of the ways we can get a SUM of 9 in power:
(1) 0 + 9 = 9 ---> 1, m^9 when multiplied, will give = m^9
(2) 1 + 8 = 9 ---> m, m^8 when multiplied, will give = m^9
(3) 2 + 7 = 9 ---> m^2, m^7 when multiplied, will give = m^9
(4) 3 + 6 = 9 ---> m^3, m^6 when multiplied, will give = m^9
(5) 4 + 5 = 9 ---> m^4, m^5 when multiplied, will give = m^9
(6) 1 + 2 + 6 = 9 ---> m, m^2, m^6 when multiplied, will give = m^9
(7) 1 + 3 + 5 = 9 ---> m, m^3, m^5 when multiplied, will give = m^9
(8) 2 + 3 + 4 = 9 ---> m^2, m^3, m^4 when multiplied, will give = m^9
These are the only combination of products of power of m which would lead to m^9. Note that only first nine terms from (1+m) to (1+m^9) would give us m^9 terms. All remaining terms will give higher powers of m and hence not of any use for us.
Since each of the m^9 term has a coefficient = 1, so adding all eight m^9 terms, we get = 8*m^9
So, coefficient of m^9 = 8
Option B (correct)
Posted from my mobile device
03 Dec 2021, 04:33
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Official Solution:
What is the coefficient of \(m^9\) if you expand the following product: \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111})\)?
A. \(7\)
B. \(8\)
C. \(9\)
D. \(10\)
E. \(11\)
When you expand \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111})\) you'll get:
\(1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9+...\).
We need to find the coefficient of \(m^9\), \(i\).
Notice that all terms where the exponent is higher that 9 (\((1 + m^{10}), \ (1 + m^{11}), \ ..., \ (1 + m^{111})\)) will not contribute to the coefficient of \(m^9\) so they can be ignored. For example, say we have: \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{9})(1 + m^{10})\).
When we expand the first nine terms we'd get:
\((1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)(1 + m^{10})\)
And when we further expand this we'd get:
\(1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9+\)
\(+ (1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)*m^{10}\).
Notice that the coefficient of \(m^9\), \(i\), remained the same (all terms of \((1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)*m^{10}\)) will have \(m\) in higher power than 9).
So, we can focus only on \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^9)\)
Now, we know that \(a^x*a^y*a^z*...=a^{x+y+z+...}\), so to get the coefficient of \(m^9\) we should find out in how many ways the exponents can sum to 9.
If we have one term we can just have \(m^9\). 1 way.
If we have two terms we can have \(m*m^8=m^9\), \(m^2*m^7=m^9\), \(m^3*m^6=m^9\), or \(m^4*m^5=m^9\). 4 ways.
If we have three terms we can have \(m*m^2*m^6=m^9\), \(m*m^3*m^5=m^9\), or \(m^2*m^3*m^4=m^9\). 3 ways.
Four or more terms will not give the exponent of 9 because the least exponent will be \(m*m^2*m^3*m^4=m^{10}\).
Total of \(1+3+4=8\) ways.
So, the coefficient of \(m^9\) is 8 For those who are interested:
\((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^9)=\)
\(1 + m + m^2 + 2 m^3 + 2 m^4 + 3 m^5 + 4 m^6 + 5 m^7 + 6 m^8 + 8 m^9 +... + m^{45}\).
Answer: B
What is the coefficient of \(m^9\) if you expand the following product: \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111})\)?
A. \(7\)
B. \(8\)
C. \(9\)
D. \(10\)
E. \(11\)
When you expand \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111})\) you'll get:
\(1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9+...\).
We need to find the coefficient of \(m^9\), \(i\).
Notice that all terms where the exponent is higher that 9 (\((1 + m^{10}), \ (1 + m^{11}), \ ..., \ (1 + m^{111})\)) will not contribute to the coefficient of \(m^9\) so they can be ignored. For example, say we have: \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{9})(1 + m^{10})\).
When we expand the first nine terms we'd get:
\((1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)(1 + m^{10})\)
And when we further expand this we'd get:
\(1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9+\)
\(+ (1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)*m^{10}\).
Notice that the coefficient of \(m^9\), \(i\), remained the same (all terms of \((1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)*m^{10}\)) will have \(m\) in higher power than 9).
So, we can focus only on \((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^9)\)
Now, we know that \(a^x*a^y*a^z*...=a^{x+y+z+...}\), so to get the coefficient of \(m^9\) we should find out in how many ways the exponents can sum to 9.
If we have one term we can just have \(m^9\). 1 way.
If we have two terms we can have \(m*m^8=m^9\), \(m^2*m^7=m^9\), \(m^3*m^6=m^9\), or \(m^4*m^5=m^9\). 4 ways.
If we have three terms we can have \(m*m^2*m^6=m^9\), \(m*m^3*m^5=m^9\), or \(m^2*m^3*m^4=m^9\). 3 ways.
Four or more terms will not give the exponent of 9 because the least exponent will be \(m*m^2*m^3*m^4=m^{10}\).
Total of \(1+3+4=8\) ways.
So, the coefficient of \(m^9\) is 8 For those who are interested:
\((1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^9)=\)
\(1 + m + m^2 + 2 m^3 + 2 m^4 + 3 m^5 + 4 m^6 + 5 m^7 + 6 m^8 + 8 m^9 +... + m^{45}\).
Answer: B
