Question for Maths Legends!
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Answers
☑ Required Answer-
Answer is 6
To put it simply, 8 is not equal to 56 and 7 is not equal to 42 and 6 is not equal to 30 and 5 is not equal to 20. If you think otherwise, then I will give you $8 and you give me back $56 and we will call it even.
In mathematics, symbols are important. Some symbols are allowed to do double duty; but they don’t do so arbitrarily.
The simplest way to state what you seem to wish to convey would be:
8 => 56, 7 => 42, 6 => 30, 5 => 20. If 3 => x, what must x be? In this case, we are using the symbol => in its lesser, but valid, sense of “giving”. Function notation is even better: f(8) = 56, etc. Best would be a sequence: 8th term is 56, 7th term is 42, etc.
If there is one symbol central to the core of mathematics, it is probably the equal sign. Give it the value it is worth; and don’t tamper with its proper usages.
Answer:
Coefficient of x⁶y³ is 672.
Step-by-step explanation:
General term of expansion (a+b)ⁿ is
\bf \: T_{r+1} = \: \: ^nC_r \: \: \large \frak{ a ^{n−r} b ^r}T
r+1
=
n
C
r
a
n−r
b
r
For (x+2y)⁹,
Putting n =9, a=x, b=2y
\begin{gathered} \bf \: T_{r+1} = \: \: ^{9} C_r (x) ^{9−r} (2y) ^r \\ \\ \bf \: T _{r+1} = \: \: ^{9} C_r (x) ^{9−r} .(y) ^r .(2) ^r\end{gathered}
T
r+1
=
9
C
r
(x)
9−r
(2y)
r
T
r+1
=
9
C
r
(x)
9−r
.(y)
r
.(2)
r
Comparing with x⁶ y³ , we get, r = 3
Therefore,
\begin{gathered} \bf \: T _{r+1} \\ \: \tt ^9C_3 (x)^9−3 .y³ .2³ \\ \tt\: \: 9! (2)³× x⁶ × y³) / (3!.6!) \\ \: \tt 672x⁶ y³\end{gathered}
T
r+1
9
C
3
(x)
9
−3.y³.2³
9!(2)³×x⁶×y³)/(3!.6!)
672x⁶y³