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

Answer:

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

Step-by-step 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

Answer 2

The sum of the given series are:

(a) 79800

(b) 150975

(c) 1560

(a)

Given:

• $1 + 2 + 3 + 4 + . . . + 396 + 397 + 398 + 399$

As we know the formula,

• $1+2+3+...+n = \frac{n(n+1)}{2}$

The sum will be:

= $\frac{399(399+1)}{2}$

= $\frac{399\times 400}{2}$

= $399\times 200$

= $79800$

(b)

Given:

• $1 + 2 + 3 + 4 + . . . + 546 + 547 + 548 + 549$

The sum will be:

= $\frac{549(549+1)}{2}$

= $\frac{549\times 550}{2}$

= $549\times 275$

= $150975$

(c)

Given:

• $2 + 4 + 6 + 8 + . . . + 72 + 74 + 76 + 78$

Thu sum will be:

= $2(1+2+3+4+...+36+37+38+39)$

= $2\times {\frac{39(39+1)}{2} }$

= $39\times 40$

= $1560$

Thus the above answers are correct.

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