Time complexity using back substitution

Question

T(n) = T(n-1) + n^4

Comments

$T(n) = T(n-1) + n^4$

$T(n) = T(n-2) + n^4 + (n-1)^4$

$T(n) = T(n-3) + n^4 + (n-1)^4  + (n-2)^4$

.

.

.

$T(n) = T(n-k) + n^4 + (n-1)^4 + (n-2)^4 + … + (n-k+1)^4$

Taking base case T(1) = 1 since it is not mentioned, $n-k = 1 => k = n-1$

$T(n) = T(1) + n^4 + (n-1)^4 + … + (2)^4$

$T(n) = 1^4 + 2^4 + 3^4 + … + n^4$

$T(n) = \frac{n(n+1)(2n+1)(3n^2+3n+1)}{30} = O(n^5)$

---

Thanks. I am a newbie in algorithms and I was struggling here. Thank you so much. :-)

---

@This_is_Nimishka no problem :)