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#
