$\frac{n(n+1)}{2}$ বর্গ

$\frac{(2n+1)n(n+1)}{6}$ বর্গ

$n(n+1)^2$ বর্গ

$\left(\frac{n(n+1)}{2}\right)^2$

$S_n = \frac{n(n + 1)}{2}$

$S_n = \left(\frac{n(n + 1)}{2}\right)^2$

$S_n = \frac{n(n + 1)(2n + 1)}{6}$

$S_n = n^2(n + 1)^2$

$\left(\frac{n(n+1)}{2}\right)^2$

$\frac{n^2(n+1)^2}{4}$

$\frac{1}{4} (n(n+1))^2$

$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$

$(r + 1)^2 - (r - 1)^2 = 4r$

$(r + 1)^2r^2 - r^2(r - 1)^2 = 4r^3$

$4S_n = (n+1)^2n^2$

$4S_n = n^2(n + 1)^2$

$36$

$64$

$27$

$16$