Question

(complete solution)
can someone help me here please thank you everyone! lovelots​

Answer 1

The expression a2+a4+a6+a8+......a20 can be written in sigma notation as:

∑ a(2n) where n = 1 to 10

To expand this summation, we can substitute in the values of n and evaluate the series:

a(21) + a(22) + a(23) + ... + a(210) = a2 + a4 + a6 + ... + a20

The expression (-1) + 2 +(-3) + 4 + (-5) ....+(-25) can be written in sigma notation as:

∑ (-1)^n*n where n = 1 to 25

To expand this summation, we can substitute in the values of n and evaluate the series:

(-1)^11 + (-1)^22 + (-1)^33 + ... + (-1)^2525 = -1 + 2 - 3 + 4 - 5 +... - 25

Note that the series (-1)^n*n is an alternating series, where the signs of the terms alternate between positive and negative. This series has no closed form solution, but we can evaluate them by adding and subtracting each term.

Answer 2

`\displaystyle \sum_(n=1)^(10) a_((2n))`

`\displaystyle \sum_(n=1)^(25) n (-1)^(n)`

Explanation:

Part I

Sigma notation means the sum of the series.

Given series:

• a₂ + a₄ + a₆ + a₈ + ... + a₂₀

Therefore:

• First term is a₂
• Second term is a₄
• Third term is a₆

So each term is a₂ₙ

Therefore, the expression in sigma notation is:

`\displaystyle \sum_(n=1)^(10)a_((2n))`

Given series:

• (-1) + 2 + (-3) + 4 + (-5) + ... + (-25)

The absolute value of each term of the series is n.

The signs of each term alternate between negative and positive.

Therefore, the expression in sigma notation is:

`\displaystyle \sum_(n=1)^(25) n (-1)^(n)`

Part II

The expansion of each summation has been given in Part I.

However, the full expansions are:

`\displaystyle \sum_(n=1)^(10) a_((2n))=a_2+a_4+a_6+a_8+a_(10)+a_(12)+a_(14)+a_(16)+a_(18)+a_(20)`

`\displaystyle \sum_(n=1)^(25) n (-1)^(n)=(-1)+2+(-3)+4+(-5)+6+(-7)+8+(-9)+10+(-11)+12+\\\phantom{wwwwwwww}(-13)+14+(-15)+16+(-17)+18+(-19)+20+(-21)+22+\\\\\phantom{wwwwwwww}(-23)+24+(-25)`

Part III

The evaluation of each series is:

`\displaystyle \sum_(n=1)^(10) a_((2n))=a_2+a_4+a_6+a_8+a_(10)+a_(12)+a_(14)+a_(16)+a_(18)+a_(20)`

`\displaystyle \sum_(n=1)^(25) n (-1)^(n)=-13`