On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 20 days?
A. 39
B. 40
C. 210
D. 400
E. 401

Arithmetic sequences and Gauss summation中文
Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

1. 2015 AMC 8 官方解析

1+3+5+⋯+[(20−1)×2+1]=1+3+5+⋯+39=(1+39)×20÷2=4001 + 3 + 5 + \cdots + [(20 - 1) \times 2 + 1] = 1 + 3 + 5 + \cdots + 39 = (1 + 39) \times 20 \div 2 = 400。

The sum of 1,3,5,⋯ ,391, 3, 5, \cdots, 39 is (1+39)(20)2=400\frac{(1 + 39)(20)}{2} = 400；也可以看成前 20 个奇数之和，即 202=40020^2 = 400。

Answer

D. 400

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1已知数列{an}\{a_{n}\}的前nn项和为SnS_{n}，首项为a1a_{1}，且44，ana_{n}，SnS_{n}成等差数列． (1)求数列{an}\{a_{n}\}的通项公式； (2)若an2=2bna_{n}^{2}={{2}^{{{b}_{n}}}}…
  2. 2已知数列{an}\{a_{n}\}为等差数列，且a1+a7+a13=πa_{1}+a_{7}+a_{13}=π，则tan⁡(a2+a12)\tan (a_{2}+a_{12})的值为()() A. 3 \sqrt {3} B. −3- \sqrt {3}…
  3. 3When Yunji added all the integers from 11 through 99, she mistakenly left out a number. Her incorrect sum turned out t…
  4. 4Lucius is counting backward by 7s. His first three numbers are 100, 93, and 86. What is his 10th number? A. 30 B. 37 C.…
  5. 5已知等差数列 {an}\{a_n\} 中，a1=3a_1 = 3，公差 d=2d = 2，求前10项和 S10S_{10}。