Write a function called max_sum that takes v, a ro vector of numbers, and n, a positive integer as inp the largest possible. In other words, if v is [1 2 3 4 5 1] and n is 3, it will find 4 5 and 4 because their suc sequences exist in v, max_sum returns the first one

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Write a function called max_sum that takes v, a row
vector of numbers, and n, a positive integer as inputs. T is
the largest possible. In other words, if v is [1 2 3 45 43 2
1] and n is 3, it will find 4 5 and 4 because their such
sequences exist in v, max_sum returns the first one. The
function returns summa, the sum as the first consecutive
ones as the second output. If the input n is larger than the
number of elements of v, the function runs:
[summa, index] = max_sum([1 2 3 4 5 4 3 2 1],3)
%3D
summa =
13
index = 4
[summa, index] = max_sum([1 2 3 4 5 4 3 2 1],2)
summa = 9
index = 4
[summa, index] = max_sum([1 2 345 4 3 2 1],1)
summa
5
index = 5
%3D
[summa, index] = max_sum([1 2 3 4 5 4 3 2 1],9)
summa = 25
index = 1
[summa, index] = max_sum([1 2 3 4 5 4 3 2 1],10)
summa =
index = -1
Transcribed Image Text:Write a function called max_sum that takes v, a row vector of numbers, and n, a positive integer as inputs. T is the largest possible. In other words, if v is [1 2 3 45 43 2 1] and n is 3, it will find 4 5 and 4 because their such sequences exist in v, max_sum returns the first one. The function returns summa, the sum as the first consecutive ones as the second output. If the input n is larger than the number of elements of v, the function runs: [summa, index] = max_sum([1 2 3 4 5 4 3 2 1],3) %3D summa = 13 index = 4 [summa, index] = max_sum([1 2 3 4 5 4 3 2 1],2) summa = 9 index = 4 [summa, index] = max_sum([1 2 345 4 3 2 1],1) summa 5 index = 5 %3D [summa, index] = max_sum([1 2 3 4 5 4 3 2 1],9) summa = 25 index = 1 [summa, index] = max_sum([1 2 3 4 5 4 3 2 1],10) summa = index = -1
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