Question

If 1cube+2cube+3cube+.......+K cube=44100 then find 1+2+3+......k

Answer 1

Answer:
$1 + 2 + 3 + ... + k = 210$

Solution:

As we know that

${1}^{3} + {2}^{3} + {3}^{3} + ... + {n}^{3} = \bigg({ \frac{n(n + 1)}{2} }\bigg)^{2} \\ \\ so \\ \\ {1}^{3} + {2}^{3} + {3}^{3} + ... + {k}^{3} = \bigg({ \frac{k(k + 1)}{2} }\bigg)^{2}$
we also know that

$1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2} \\ \\ so \\ \\ 1 + 2 + 3 + ... + k= \frac{k(k + 1)}{2}$
it is here given that
${1}^{3} + {2}^{3} + {3}^{3} + ... + {k}^{3} = \bigg({ \frac{k(k + 1)}{2} }\bigg)^{2} \\ \\ \\\bigg({ \frac{k(k + 1)}{2} }\bigg)^{2} = 44100 \\ \\ \frac{k(k + 1)}{2} = \sqrt{44100} \\ \\ \frac{k(k + 1)}{2} = 210 \\ \\ \\ hence \\ \\ 1 + 2 + 3 + ... + k = 210$

Answer 2

Answer:

Step-by-step explanation: