Question #0b5ad

1 Answer
Oct 21, 2017

n=4

Explanation:

Recall that a summation, symbolized with the symbol sum, represents a running sum (addition) of terms in a sequence, beginning with some variable value (shown under the sum, while ending with some variable value (shown above the sum.

The expression sum_{k=1}^n (2k-1), then, means you do the following process:

  • Evaluate 2k-1 by substituting k = 1
  • Add to that what you get if you evaluate 2k-1 by substituting k = 2.
  • Add to that what you get if you evalute 2k-1 by substituting k = 3.
  • Keep adding terms until you eventually evaluate 2k-1 by substituting k = n (the last value for k).

Mathematically, we'd write this:

sum_{k=1}^n (2k-1) =

(2(1)-1) + (2(2)-1) + ... + (2(n-2)-1) + (2(n-1)-1) + (2n-1)

However, we are then subtracting another summation, of the exact same expression, but this time running from k=1 to k=n-1. That looks like this:

sum_{k=1}^(n-1) (2k-1) =

(2(1)-1) + (2(2)-1) + ... + (2(n-2)-1) + (2(n-1)-1)

If you are observant, you'll notice that sum_(k=1)^n (2k-1) contains every single term that you see in sum_(k=1)^(n-1) (2k-1), with the addition of one final term (2n-1) being added to it. So:

sum_{k=1}^n (2k-1) - sum_{k=1}^(n-1) (2k-1)

=color(white)("aaaa"){: ((2(1)-1) + (2(2)-1) + ... + (2(n-2)-1) + (2(n-1)-1) + (2n-1)),(color(white)("aaaaaaaaaaaaaaaaaaa")-),((2(1)-1) + (2(2)-1) + ... + (2(n-2)-1) + (2(n-1)-1)):}

= (2n-1) = 7 :. 2n = 8 => n = 4

Verification

sum_{k=1}^4 (2k-1) - sum_{k=1}^3 (2k-1)

= (2*1-1) + (2*2-1) + (2*3-1) + (2*4-1) - [(2*1-1) + (2*2-1) + (2*3-1) ]

= 1 + 3 + 5 + 7 - [1 + 3 + 5] = 16 - 9 = 7