Question 42 of 51advanced🔍 AnalyzeLong Answer5 marks

Find all possible ways of expressing 100100 as the sum of consecutive natural numbers.

Correct Answer

100

18 + 19 + 20 + 21 + 22

9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Exercise: END-OF-CHAPTER EXERCISES | Q: 6 | (Page 22)
For More Understanding

Explanation

The textbook introduces the formula for the sum of the first nn natural numbers, Sn=n(n+1)2S_n = \frac{n(n+1)}{2}, and shows how to find the sum of consecutive numbers by subtracting sums. Using this, we set up an equation for the sum of consecutive numbers from a+1a+1 to bb equaling 100100.

Factoring leads to (b−a)(b+a+1)=200(b-a)(b+a+1) = 200. Since the two factors must have different parities, we test the odd factors of 200200 (11, 55, 2525) to find valid natural number sequences.

Solution Steps

  1. Step 1: Use the formula for the sum of consecutive natural numbers from a+1a+1 to bb:

    Sb−Sa=b(b+1)2−a(a+1)2=100S_b - S_a = \frac{b(b+1)}{2} - \frac{a(a+1)}{2} = 100
  2. Step 2: Multiply by 2 and rearrange:

    b(b+1)−a(a+1)=200b(b+1) - a(a+1) = 200

    which simplifies to

    (b−a)(b+a+1)=200(b-a)(b+a+1) = 200
  3. Step 3: Let x=b−ax = b-a and y=b+a+1y = b+a+1. Since y−x=2a+1y - x = 2a + 1 is odd, xx and yy must have different parities (one odd, one even).

  4. Step 4: The odd factors of 200200 are 11, 55, and 2525. The valid pairs (x,y)(x, y) are (1,200)(1, 200), (5,40)(5, 40), and (8,25)(8, 25).

  5. Step 5: For (x,y)=(1,200)(x, y) = (1, 200), b=100b=100 and a=99a=99, giving the sequence:

    100100
  6. Step 6: For (x,y)=(5,40)(x, y) = (5, 40), b=22b=22 and a=17a=17, giving the sequence:

    18+19+20+21+2218 + 19 + 20 + 21 + 22
  7. Step 7: For (x,y)=(8,25)(x, y) = (8, 25), b=16b=16 and a=8a=8, giving the sequence:

    9+10+11+12+13+14+15+169 + 10 + 11 + 12 + 13 + 14 + 15 + 16