Find the sum of ( 1- 1/n) + (1 - 2/n) + ( 1 - 3/n)...upto n terms

(n-1) / 2

Step-by-step explanation:

We have that the sum has the form of:

Sn = n / 2 [2 * a + (n-1) * d]

we compute the terms of a and d

a = 1 - 1 / n

a = (n - 1) / n

, that is, the first term, in the case d is the subtraction between the second and the first term, thus:

d = 1 - 2 / n - (1 - 1 / n)

d = 1 - 2 / n - 1 + 1 / n

d = - 1 / n

now if replacing these values:

Sn = n / 2 * [2 * (n - 1) / n + (n-1) * (- 1 / n)

Sn = n / 2 * [2 * (n-1) - (n-1)] / n

Sn = 1/2 * (n-1)

Therefore, the value of that sum is (n-1) / 2

when studying polynomials, you often hear the terms zeros, roots, factors and x x the multiplicity of a zero is important because it tells us how the graph of the let's look at the graph of a function that has the same zeros

step-by-step explanation:

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions