2016 math

Algebra Level 3

$\begin{array} {} (A): & 2016 = 1 + 2 + 3 +\cdots+ 62 \\ (B): & 2016 = \sqrt{1^3 + 2^3 +\cdots+ 63^3} \\ (C): & 2016 = 2^{10} + 2^9 +\cdots+ 2^4 \\ (D): & 2016 = 1^2 - 2^2 + 3^2 - 4^2 +\cdots-62^2 + 63^2 \end{array} $

Consider the statements above.

Then,

Both

(A)

and

(D)

are true Both

(B)

and

(C)

are true Both

(B)

and

(D)

are true Only

(D)

is true and others are false.

1 solution

Chew-Seong Cheong
May 6, 2016

$\begin{aligned} A & = 1 + 2 + 3 +\cdots + 62 \\ & = \sum_{n=1}^{62} n = \frac{62(63)}{2} = 1953 \color{#D61F06} {\ne 2016} \\ B & = \sqrt{1^3 + 2^3 + 3^3 +\cdots + 63^3} \\ & = \sqrt{\sum_{n=1}^{63} n^3} = \sqrt{\left(\frac{63(64)}{2}\right)^2} \color{#3D99F6}{= 2016} \\ C & = 2^{10}+2^9+2^8+\cdots+2^4 \\ & = 2^4 \sum_{n=0}^6 2^n = \frac{2^4\left(2^7-1\right)}{2-1} = 2032 \color{#D61F06} {\ne 2016} \\ D & = 1^2 - 2^2 + 3^2 - 4^2 +\cdots - 62^2 + 63^2 \\ & = \sum_{n=1}^{63} n^2 - 2(2^2) \sum_{n=1}^{31} n^2 = 85344 - 8(10416) \color{#3D99F6}{= 2016} \end{aligned}$

Therefore, $\boxed{\text{both (B) and (D) are true}.}$

Nice solution, sir.

akash patalwanshi - 5 years, 1 month ago

The $D$ can also be evaluated more easily by using $a^2-b^2= (a+b)(a-b)$ which can be written as $-1-2-3-4-...-61-62+63^2 = 63^2-31\times 63 = 63\times 32 = 2016$

Anand Chitrao - 5 years, 1 month ago

You are right. I have edited your comment into LaTex. You can check to see how it is done by clicking Edit.

Chew-Seong Cheong - 5 years, 1 month ago

$C = 2032$ and not $2031$

Krutarth Patel - 3 months, 4 weeks ago

Thank you.

Chew-Seong Cheong - 3 months, 4 weeks ago

2016 math

1 solution

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