How to find missing angles
1 answer:
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The coefficients of the binomial expansion 
, where n is the row number, is given in the Pascal's triangle shown below.
First, to find the <span>coefficient of the fourth term of (x+2)^5 we look at row 5, term 4. The coefficient there is 10.
But, we must also remember that the term 2 also is taken to a certain power here. Mainly , for each term, the power of 2 is as follows:
2^0, 2^1, 2^2, 2^3=8.
So, in total we have: 10*8=80.
Second, to find </span><span> the coefficient of the third term of (3x-1)^5 we again go to the row 5, this time term 3 and we have 10 there. Now we must check how each of (3x) and 1 expand, now being careful about the sign as well.
we have:
(3x)^5 (1) -(3x)^4 (1) (3x)^3(1)=27x^3.
Thus, the coefficient of the third term is 27*10=270.
Third, we want to find </span><span>the coefficient of the third term of (a+5b^2)^4. We look at row 4, term 3. There we have 6.
The terms a and 5b^2 are as follows:
a^4 (5b^2)^0 </span> a^3 (5b^2)^1 a^2 (5b^2)^2=25a^2b^4
Thus, the coefficient is 25*6=150.
Answer:
80; 270; 150
First, to find the <span>coefficient of the fourth term of (x+2)^5 we look at row 5, term 4. The coefficient there is 10.
But, we must also remember that the term 2 also is taken to a certain power here. Mainly , for each term, the power of 2 is as follows:
2^0, 2^1, 2^2, 2^3=8.
So, in total we have: 10*8=80.
Second, to find </span><span> the coefficient of the third term of (3x-1)^5 we again go to the row 5, this time term 3 and we have 10 there. Now we must check how each of (3x) and 1 expand, now being careful about the sign as well.
we have:
(3x)^5 (1) -(3x)^4 (1) (3x)^3(1)=27x^3.
Thus, the coefficient of the third term is 27*10=270.
Third, we want to find </span><span>the coefficient of the third term of (a+5b^2)^4. We look at row 4, term 3. There we have 6.
The terms a and 5b^2 are as follows:
a^4 (5b^2)^0 </span> a^3 (5b^2)^1 a^2 (5b^2)^2=25a^2b^4
Thus, the coefficient is 25*6=150.
Answer:
80; 270; 150
Answer:
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