Question

Use Polya's four-step problem-solving strategy and the problem-solving procedures presented in this lesson to solve the following exercise. Find the following sums without using a calculator or a formula. Hint: Apply the procedure used by Gauss. (See the Math Matters on page 31.) (a) 1 + 2 + 3 + 4 + . . . + 396 + 397 + 398 + 399

(b) 1 + 2 + 3 + 4 + . . . + 546 + 547 + 548 + 549

(c) 2 + 4 + 6 + 8 + . . . + 72 + 74 + 76 + 78

Answer 1

a) 1+2+3+4+...+396+397+398+399=79800

b) 1+2+3+4+...+546+547+548+549=150975

c) 2+4+6+8+...+72+74+76+78=1560

Explanation:

We know that a summation formula for the first n natural numbers:

1+2+3+...+(n-2)+(n-1)+n=\frac{n(n+1)}{2}

We use the formula, we get

a) 1+2+3+4+...+396+397+398+399=\frac{399·(399+1)}{2}=\frac{399· 400}{2}=399· 200=79800

b) 1+2+3+4+...+546+547+548+549=\frac{549·(549+1)}{2}=\frac{549· 550}{2}=549· 275=150975

c)2+4+6+8+...+72+74+76+78=S / ( :2)

1+2+3+4+...+36+37+38+39=S/2

\frac{39·(39+1)}{2}=S/2

\frac{39·40}{2}=S/2

39·40=S

1560=S

Therefore, we get

2+4+6+8+...+72+74+76+78=1560