Question

please solve please

Answer 1

2^3+4^3+6^3...........
 
WE CAN TAKE 2^3 COMMON FROM THIS
SO 2^3+2^3.2^3+2^3.3^3......
       2^3(1^3+2^3+3^3.....)
       IF WE n=3 
          then   2^3(1+2^3+3^3)=8(1+8+27)=288
   
nd cross check the options by option 1) 244
                                                            2) 288
so the correct ans is option 2

Answer 2

$Given, 2^3 + 4^3 + 6^3 ........$

= > 8 + 64 + 216 ..........

$= \ \textgreater \ 8(1^3 + 2^3 + 3^3 .......)$

We know that sum of the cubes of the first n natural numbers = n^2(n+1)^2/4.

$= \ \textgreater \ 8 * \frac{n^2(n + 1)^2}{4}$

$= \ \textgreater \ 2n^2(n+1)^2$



Hope this helps!