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Answer:
![(a+b)^n ={n \choose 0}a^{(n)}b^{(0)} + {n \choose 1}a^{(n-1)}b^{(1)} + {n \choose 2}a^{(n-2)}b^{(2)} + ..... +{n \choose n}a^{(0)}b^{(n)}](https://tex.z-dn.net/?f=%28a%2Bb%29%5En%20%20%3D%7Bn%20%5Cchoose%200%7Da%5E%7B%28n%29%7Db%5E%7B%280%29%7D%20%2B%20%7Bn%20%5Cchoose%201%7Da%5E%7B%28n-1%29%7Db%5E%7B%281%29%7D%20%2B%20%7Bn%20%5Cchoose%202%7Da%5E%7B%28n-2%29%7Db%5E%7B%282%29%7D%20%2B%20.....%20%20%2B%7Bn%20%5Cchoose%20n%7Da%5E%7B%280%29%7Db%5E%7B%28n%29%7D)
Step-by-step explanation:
The Given question is INCOMPLETE as the statements are not provided.
Now, let us try and solve the given expression here:
The given expression is: ![(a +b)^n, n > 0](https://tex.z-dn.net/?f=%28a%20%2Bb%29%5En%2C%20n%20%3E%200)
Now, the BINOMIAL EXPANSION is the expansion which describes the algebraic expansion of powers of a binomial.
Here, ![(a+b)^n = \sum_{k=0}^{n}{n \choose k}a^{(n-k)}b^{(k)}](https://tex.z-dn.net/?f=%28a%2Bb%29%5En%20%20%3D%20%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%7Bn%20%5Cchoose%20k%7Da%5E%7B%28n-k%29%7Db%5E%7B%28k%29%7D)
or, on simplification, the terms of the expansion are:
![(a+b)^n ={n \choose 0}a^{(n)}b^{(0)} + {n \choose 1}a^{(n-1)}b^{(1)} + {n \choose 2}a^{(n-2)}b^{(2)} + ..... +{n \choose n}a^{(0)}b^{(n)}](https://tex.z-dn.net/?f=%28a%2Bb%29%5En%20%20%3D%7Bn%20%5Cchoose%200%7Da%5E%7B%28n%29%7Db%5E%7B%280%29%7D%20%2B%20%7Bn%20%5Cchoose%201%7Da%5E%7B%28n-1%29%7Db%5E%7B%281%29%7D%20%2B%20%7Bn%20%5Cchoose%202%7Da%5E%7B%28n-2%29%7Db%5E%7B%282%29%7D%20%2B%20.....%20%20%2B%7Bn%20%5Cchoose%20n%7Da%5E%7B%280%29%7Db%5E%7B%28n%29%7D)
The above statement holds for each n > 0
Hence, the complete expansion for the given expression is given as above.
It could be The 4.0 cm would be the base and 2.8cm would be the length upwards
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