try this without a calculator

Algebra Level 2

Determine 1+3+5+7+9+...+221

The answer is 12321.

3 solutions

Bk Lim
Dec 18, 2014

There are 111 numbers with average 111.

111 x 111 =12321

Abdur Rehman Zahid
Dec 7, 2014

First take a look at the partial sums $\color{#20A900}{1+3=\left(\frac{1+3}{2}\right)^2=2^2=4\\1+3+5=\left(\frac{1+5}{2}\right)^2=3^2=9\\1+3+5+7=\left(\frac{1+7}{2}\right)^2=4^2=16}$ This suggest to us that the sum of $\color{#3D99F6}{1+3+5+\dotsm+(2n-1)=\left(\frac{1+2n-1}{2}\right)^2}$ .So the sum of the arithmetic progression in question is $\color{#D61F06}{\left(\frac{1+221}{2}\right)^2=111^2=\boxed{12321}}$

That sum is actually an identity for $n\in \mathbb{Z^+}$ which can easily be proved using arithmetic progression formulas. Furthermore, there's a trick to evaluating the value of $(111)^2$ without using a calculator.

$(\overline{111\ldots n \text{ times }})^2 = \overline{123\ldots (n-1)n(n-1)\ldots 321}, n\in [1,9] \land n\in \mathbb{N}$

That gives us $(111)^2=\boxed{12321}$

Note that in the trick I depicted, the values are not in multiplicative form but rather are the digits of the number which the overline is used to represent. For example, if $P=1,Q=5,R=8,S=6$ , then $\overline{PQRS}=1586$

Prasun Biswas - 6 years, 6 months ago

Rodette Batacandolo
Dec 5, 2014

for the series 1+3+5+7+9+... with n terms, the summation will be equal to n^2 Solving for n, by arithmetic progression

221=1+(n-1)2 ; n = 111 ; the answer should be 111^2 = 12321

try this without a calculator

The answer is 12321.

3 solutions

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