Using arithmetic with arrays

Your for loop is malformed. You need to use < instead of <=, use n instead of 5, and use ; instead of ,.

Your assignment of multiple is backwards. The value on the right-side of the = operator is assigned to the variable on the left-side of =. You are trying to assign the value of multiple (which is uninitialized) to a dynamically computed value that has no explicit variable of its own. You should be assigning the computed value to multiple instead.

Also, you didn't follow the "this number [array dimensions] should be passed as a parameter to the function" requirement of your instructions.

Try this:

#include <iostream>   
using namespace std; 

const int n = 5;

void function(const int *array1, const int *array2, const int size);

int main()
{
    int array1[n] = { 5,  3, -4, 2, 8};
    int array2[n] = {-7, -9,  5, 2, 9};

    function(array1, array2, n);

    return 0;
}

void function(const int *array1, const int *array2, const int size)
{
    int multiple = 0;

    for(int i = 0; i < size; i++)
    {
        multiple += (array1[i] * array2[i]);
    }

    cout << multiple << endl;
}

Syntax error is where you are using the for loop in this fashion:

for(int i=0;i<=5,i++)

Use this instead:

for(int i=0; i <= 5; i++)

The function can assume that the vectors have the same dimension (n, say), but this number should be passed as a parameter to the function.

This function declaration

void function(int array[n],int array2[n]);

does not include a parameter that specifies the dimension of the arrays.

The above declaration is equivalent to

void function(int *array,int *array2);

because arrays passed by value are implicitly converted to pointers to their first elements.

There are a typo and a bug in this for statement

for(int i=0; i <=5, i++)
             ^^^^^^

There shall be

for ( int i=0; i < n; i++)

The variable multiple

int multiple;

is not initialized and this assignment

(array[i]*array2[i]) + (array[i+1]*array2[i+1]) = multiple;

does not have sense and has nothing common with the condition

nuv = u1v1 + u2v2 + u3v3 + u4v4 +………… + unvn = 0

It seems what you mean is the following

#include <iostream>

bool function( const int array[], const int array2[], size_t n ) 
{

    long long int product = 0;

    for ( size_t i = 0; i < n; i++)
    {
        product += array[i] * array2[i];
    }

    return product == 0;
}

int main()
{
    const size_t N = 5;
    int array[N]  = { 5, 3 , -4, 2, 8 };
    int array2[N] = { -7, -9, 5, 2, 9 };

    std::cout << "array and array2 are "
        << (function(array, array2, N) ? "" : "not ")
        << "at right angles"
        << std::endl;

    return 0;
}

These arrays

    int array[N]  = { 5, 3 , -4, 2, 8 };
    int array2[N] = { -7, -9, 5, 2, 9 };

are not a right angles,

But these arrays

    int array[N]  = { 5, 3 , -4, 1, 9 };
    int array2[N] = { -7, -9, 5, 1, 9 };

are at right angles. Try them.

Alternative: try the C++ way. Use std::array that knows it's length. Use algorithms provided by the standard library like std::inner_product.

#include <iostream>
#include <algorithm>
#include <array>
#include <numeric>

int main()
{
  using arr_t = std::array<int,5>;

  arr_t arr1 = {5, 3 , -4, 2, 8};
  arr_t arr2 = {-7, -9, 5, 2, 9};

  int mult = std::inner_product( begin(arr1), end(arr1), begin(arr2), 0,
                 std::plus<>(), std::multiplies<>() );

  std::cerr << mult << "\n";
}