An Interesting Sum!

Algebra Level 2

$1\cdot 2^2+2 \cdot 3^2+3 \cdot 4^2+\cdots+2019 \cdot 2020^2$

Evaluate the sum above.

\sum_{n=1}^{2020}\frac{ n^3-n^ 2}{2}

\sum_{n=1}^{2020} n^3-\sum_ {n=1}^{2020} n^2

\sum_{n=1}^{2020} n^2

\sum_{n=1}^{2020} n^3

\sum_{n=1}^{2020} n

2 solutions

A Former Brilliant Member
Jul 25, 2020

Adding $0.1^2$ to the series, which doesn't change the sum, we see that the $n'^{th}$ term of the series is

$n^2(n-1)=n^3-n^2$

Hence option $2$ is the correct option .

Thank you for sharing your solution.

Hana Wehbi - 10 months, 3 weeks ago

Chew-Seong Cheong
Jul 26, 2020

$\begin{aligned} S & = 1 \cdot 2^2 + 2 \cdot 3^2 + 3 \cdot 4^2 + \cdots + 2019 \cdot 2020^2 \\ & = \sum_\blue{n=2}^{2020} (n-1) n^2 \\ & = \underbrace{\cancel{(1-1)1^2}^{=0}}_\red{n=1} + \sum_\blue{n=2}^{2020} (n-1) n^2 \\ & = \sum_\red{n=1}^{2020} (n-1) n^2 \\ & = \boxed{\sum_{n=1}^{2020} n^3 - \sum_{n=1}^{2020} n^2} \end{aligned}$

@Hana Wehbi , sorry but I remember telling you to use \cdot (center dot) $\cdot$ for multiplying sign and not the decimal point. Then you don't need to explain it (I have removed the explanation). I hope that you will memorize this as you memorize Python coding.

It is unnecessary to have $\cdots 2019 \cdot 2020^2 \cancel{ = \ ?}$ . It is necessary to have $n=1$ and brackets are unnecessary in $\displaystyle \sum_{n=1}^{2020} \cancel (n^3 \cancel )$ .

Chew-Seong Cheong - 10 months, 3 weeks ago

I will and thank you for reminding me. It wasn’t intentional but it didn’t come to my mind, my apologies.

Hana Wehbi - 10 months, 2 weeks ago

An Interesting Sum!

2 solutions

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