If f(x) is a function such that f(x+y) = f(x)+ f(y) and f(1) = 7 then \( \sum_{r=1}^{n}\) f(r) =
If f(x) is a function such that f(x+y) = f(x)+ f(y) and f(1) = 7 then \( \sum_{r=1}^{n}\) f(r) =
\(\frac{7n}{2}\)
\(\frac{7(n+1)}{2}\)
7n(n+1)
\(\frac{7n(n+1)}{2}\)
The Correct Option is D
Solution and Explanation
To solve the problem, we are given a function \( f(x) \) such that:
- \( f(x + y) = f(x) + f(y) \)
- \( f(1) = 7 \)
1. Use the Property of the Function:
The functional equation \( f(x + y) = f(x) + f(y) \) is the definition of a linear function. Specifically, a function of the form \( f(x) = kx \) satisfies this equation for all real numbers.
2. Determine the Constant:
We are given \( f(1) = 7 \). If \( f(x) = kx \), then:
\[ f(1) = k \cdot 1 = 7 \Rightarrow k = 7 \] So, the function is \( f(x) = 7x \)
3. Evaluate the Summation:
We are to find: \[ \sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r = 7 \sum_{r=1}^{n} r \] We know: \[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \] Therefore: \[ \sum_{r=1}^{n} f(r) = 7 \cdot \frac{n(n+1)}{2} = \frac{7n(n+1)}{2} \]
Final Answer:
The value of \( \sum_{r=1}^{n} f(r) \) is \( \frac{7n(n+1)}{2} \)
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Concepts Used:
Algebra of Complex Numbers
Algebra of complex numbers
1. Addition of two complex numbers:
Consider z1 and z2 are two complex numbers.
For example, z1 = 3+4i and z2 = 4+3i
Here a=3, b=4, c=4, d=3
∴z1+ z2 = (a+c)+(b+d)i
⇒z1 + z2 = (3+4)+(4+3)i
⇒z1 + z2 = 7+7i
Properties of addition of complex numbers
- Closure law: While adding two complex numbers the resulting number is also a complex number.
- Commutative law: For the complex numbers z1 and z2 , the commutation can be z1+ z2 = z2+z1
- Associative law: While considering three complex numbers, (z1+ z2) + z?3 = z1 + (z2 + z3)
- Additive identity: An Additive identity is nothing but zero complex numbers that go as 0+i0. For every complex number z, z+0 = z.
- Additive inverse: Every complex number has an additive inverse denoted as -z.
2. Difference between two complex numbers
It is similar to the addition of complex numbers, such that, z1 - z2 = z1 + ( -z2)
For example: (5+3i) - (2+1i) = (5-2) + (-2-1i) = 3 - 3i
3. Multiplication of complex numbers
Considering the same value of z1 and z2 , the product of the complex numbers are
z1 * z2 = (ac-bd) + (ad+bc) i
For example: (5+6i) (2+3i) = (5×2) + (6×3)i = 10+18i
Properties of Multiplication of complex numbers
Note: The properties of multiplication of complex numbers are similar to the properties we discussed in addition to complex numbers.
- Closure law: When two complex numbers are multiplied the result is also a complex number.
- Commutative law: z1* z2 = z2 * z1
Associative law: Considering three complex numbers, (z1 z2) z3 = z1 (z2 z3)
- Multiplicative identity: 1+0i is always denoted as 1. This is multiplicative identity. This means that z.1 = z for every complex number z.
- Distributive law: Considering three complex numbers, z1 (z2 + z3) =z1 z2 + z1 z3 and (z1+ z2) z3 = z1 z2 + z2 z3.
Read More: Complex Numbers and Quadratic Equations
4. Division of complex numbers
If z1 / z2 of a complex number is asked, simplify it as z1 (1/z2 )
For example: z1 = 4+2i and z2 = 2 - i
z1 / z2 =(4+2i)×1/(2 - i) = (4+i2)(2/(2²+(-1)² ) + i (-1)/(2²+(-1)² ))
=(4+i2) ((2+i)/5) = 1/5 [8+4i + 2(-1)+1] = 1/5 [8-2+1+41] = 1/5 [7+4i]