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25 Oct 2017, 07:42
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
55% (02:08) correct
45%
(02:00)
wrong
based on 187
sessions
History
Date
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Not Attempted Yet
If \((a+b)^5=a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5\). What is the value of x+y+z+t?
A. 7
B. 15
C. 20
D. 30
E. 35
A. 7
B. 15
C. 20
D. 30
E. 35
25 Oct 2017, 08:25
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div10
We can solve the problem by using the pascals triangle - The coefficients(the numbers in front of each term) follow a pattern
1
\((a+b)^0 = 1\)
1 1
\((a+b)^1 = a + b\)
1 2 1
\((a+b)^2 =a^2 + 2ab +b^2\)
1 3 3 1
\((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
1 4 6 4 1
\((a + b)4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b4\)
1 5 10 10 5 1
\((a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)
Therefore, Value of x+y+z+t = \(5+10+10+5 = 30\)(Option D)
25 Oct 2017, 09:23
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div10
Hi...
any expansion is ..
\((a+b)^n = nC0a^nb+nC1a^{n-1}b^1........+nCnb^n\)
and when you add the coefficient \(nC0+nC1+...nCn=2^n\)
here n is 5..
\((a+b)^5=a^5+xa^4b+ya^3b^2+za^2b^3+tab^4+b^5\)
so addition of coefficients = \(2^5=1+x+y+z+t+1......32=2+x+y+z+t\)
so \(x+y+z+t =32-2=30\)
D
