A question of degree

we all know that a sequence with a constant common difference, the general term is a one degree polynomial and the sum is $\frac{n(n+1)}{2}$ and it is a 2 degree polynomial. and the general term of a sequence with an increasing common difference is a two degree polynomial. the sum is $\frac{n(n+1)(2n+1)}{6}$ and it is a 3 degree polynomial. so we can conclude that the degree of the polynomial sum of a 23 degree polynomial general term is just by adding 1 degree to it. so it is 24.

It is given that f(x) is a polynomial.

Let $\displaystyle f(x)=\sum_{i=0}^p a_ix_i$ where $a_p \neq 0$ .

From the question, we have, $f(x)-f(x-1)=x^{23}$

Substituting $f(x)$ and $f(x-1)$ , we notice that the highest power of $x$ remaining on LHS is $p-1$ . Notice that $a_p x^p$ cancels out.

Comparing both the sides, we find that $p-1 =23 \Rightarrow p=24$ .

Therefore, the degree of $f(x)$ is 24.

Recollecting sum of N natural numbers = N ( N + 1 ) / 2..............................

.Degree 1 higher than that of what we are summing

Sum of the series ( N ^ 2 ) where N starts from 1 is N ( N +1 ) ( 2 N + 1 ) / 6........................

Degree 1 higher than that of what we are summing

Sum of the series ( N ^ 3 ) where N starts from 1 is N ( N +1 ) N ( N + 1 ) / 4........................

Degree 1 higher than that of what we are summing

Thus here too the degree should be one higher than 23 which is 24