Question

5. Determine the sum of 2+4+6+8+...+250. (Hint: 2(1+2+3+4) = 2+4+6+8)

Answer 1

Answer:

The sum of 2+4+6+8+...+250 will be: 22650

Step-by-step explanation:

Given

2+4+6+8+...+250

2(1+2+3+4+...+150)....[A]

Considering the sequence

1+2+3+4+...+150

Lets calculate the sum of first 150 terms

$\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n$

$2-1=1,\:\quad \:3-2=1$

$\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$d=1$

So the sequence is Arithmetic.

as

$a_1=1$

and

$d=1$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}$

$a_n=a_1+\left(n-1\right)d$

$a_n=\left(n-1\right)+1$

$a_n=n$

$\mathrm{Arithmetic\:sequence\:sum\:formula:}$

$n\left(a_1+\frac{d\left(n-1\right)}{2}\right)$

$\mathrm{Plug\:in\:the\:values:}$

$n=150,\:\spacea_1=1,\:\spaced=1$

$=150\left(1+\frac{1\cdot \left(150-1\right)}{2}\right)$

$=150\left(\frac{149}{2}+1\right)$

$=150\cdot \frac{151}{2}$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}$

$=\frac{151\cdot \:150}{2}$

$=\frac{22650}{2}$

$=11325$

So the sum of first 150 terms of the arithmetic

sequence 1+2+3+4+...+150 is: 11325

Now, according to expression [A], multiply 11325 by 2 to determine the sum of 2+4+6+8+...+250.

As

1+2+3+4+...+150 = 11325

Thus

2(1+2+3+4+...+150) = 2(11325) = 22650

Therefore, the sum of 2+4+6+8+...+250 will be: 22650