Easy INTEGER +4 / -1 PYQ · JEE Mains 2020

The sum $\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)}$ is :

Answer (integer) 1540

Solution

$\sum\limits_{k = 1}^{20} {\left( {1 + 2 + 3 + ... + k} \right)}$ <br><br>= $\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}}$ <br><br>= $$\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $$ <br><br>= $${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 \over 2} \times {{20 \times 21} \over 2}$$ <br><br>= 1540

About this question

Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression

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