Statement-1: The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)++(361+380+400) is 8000.

Question

Statement-1: The sum of the series $1+(1+2+4)+(4+6+9)+(9+12+16)+\cdots+(361+380+400)$ is 8000.

Statement-2: $\displaystyle\sum_{k=1}^{n}\big(k^3-(k-1)^3\big)=n^3$, for any natural number $n$.

Accepted Answer

$n$th  term of the given series

$$ = {T_n} = {\left( {n - 1} \right)^2} + \left( {n - 1} \right)n + {n^2}$$

$$ = {{\left( {{{\left( {n - 1} \right)}^3} - {n^3}} \right)} \over {\left( {n - 1} \right) - n}}$$

$$ = {n^3} - {\left( {n - 1} \right)^3}$$

$$ \Rightarrow {S_n} = \sum\limits_{k = 1}^n {\left[ {{k^3} - {{\left( {k - 1} \right)}^3}} \right]} $$

$$ \Rightarrow 8000 = {n^3}$$

$$ \Rightarrow n = 20\,\,$$  which is a natural number.

Now, put $n = 1,2,3,.....20$

${T_1} = {1^3} - {0^3}$

${T_2} = {2^3} - {1^3}$

.

${T_{20}} = {20^3} - {19^3}$

Now, $${T_1} + {T_2} + ..... + {T_{20}} = {S_{20}}$$

$$ \Rightarrow {S_{20}} = {20^3} - {0^3} = 8000$$

Hence, both the given statements are true and statement $2$ supports statement $1.$

Solutions · 1