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3 years ago

Help me pleaseeeeeeeee

Mathematics

1 answer:

romanna [79]3 years ago

8 0

Answer:

D. 7x - 1

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms/Coefficients
Functions
Function Notation

Step-by-step explanation:

Step 1: Define

Identify

f(x) = 5x - 2

g(x) = 2x + 1

Step 2: Find

Substitute in functions: (f + g)(x) = 5x - 2 + 2x + 1
Combine like terms: (f + g)(x) = 7x - 1

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3 years ago

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We have

$\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)$

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so that

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You might already know that

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

so from these formulas we get

$\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n$

$\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4$

$\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}$

If you don't know the formulas mentioned above:

The first one should be obvious; if you add $n$ copies of 1 together, you end up with $n$ .
The second one is easily derived: If $S=1+2+3+\cdots+n$ , then $S=n+(n-1)+(n-2)+\cdots+1$ , so that $2S=n(n+1)$ or $S=\dfrac{n(n+1)}2$ .
The third can be derived using a similar strategy to the one used here. Consider the expression $(n+1)^3-n^3=3n^2+3n+1$ , and so on.

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