3 years ago

Find e and d3e+d=8 2e+d=8

Mathematics

2 answers:

Westkost [7]3 years ago

7 0

Answer:

Step-by-step explanation:

son4ous [18]3 years ago

5 0

3e+d=8
2e+d=8
Multiply the second equation by -1
You get
-3e-d=-8
Use eliminating methods
-3e-d=-8
2e+d=8
We get
-e=0
So e=0
Find e by substitute it in one of the equation
2*0+d=8
D
d=8

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First show the statement holds for $n=1$ . The left hand side is just 8, and the right hand side is $4(1)(1+1)=8$ , so it's true.

Assume the statement holds for $n=k$ , i.e.

$8+16+\cdots+8(k-1)+8k=4k(k+1)$

and use this to show it holds for $n=k+1$ , i.e.

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By the assumption above, you have

$\underbrace{8+16+\cdots+8(k-1)+8k}_{n=k}+8(k+1)=4k(k+1)+8(k+1)=4(k+1)(k+2)$

so the statement is true for all $n\ge1$ .