Undetermined Coefficients (Part 2)

$\Large \sum_{n=2}^{99}\left((n-1)n(n+1)\right)$ $\large =\sum_{n=2}^{99}\left(\color{#20A900}{n^3}-\color{#D61F06}{n}\right)$ $\small{\left(\color{#20A900}{\sum n^3=\left(\dfrac{n(n+1)}{2}\right)^2},~\color{#D61F06}{\sum n=\left(\dfrac{n(n+1)}{2}\right)}\right) }$ $\large =\dfrac{n(n+1)}{2}\left(\dfrac{n(n+1)}{2}-1\right)$ $\large =\dfrac{(n-1)n(n+1)(n+2)}{4}|_{n=99}$ $\huge =\boxed{24497550}$

The required sum can be expressed as : $y=\sum_{n=1}^{98}n(n+1)(n+2)$

We can express it as difference of two consecutive terms(so that we can solve it by method of differences) by multiplying and dividing it with difference of next term after $(n+2)$ which will continue the same pattern and term before the term $n$ that will continue the same pattern:

$y=\frac{1}{4}\sum_{n=1}^{98}n(n+1)(n+2)[n+3-(n-1)]$

$\Rightarrow y=\frac{1}{4}\sum_{n=1}^{98}n(n+1)(n+2)(n+3)-(n-1)(n)(n+1)(n+2)$

$\Rightarrow \quad 4y\quad =$ $\quad \\ \quad \cancel{1.2.3.4}\quad -\quad 0.1.2.3\\ +\cancel{2.3.4.5}\quad -\quad \cancel{1.2.3.4}\\ +\cancel{3.4.5.6}\quad -\quad \cancel{2.3.4.5}\\ ...\\ ...\\ +98.99.100.101\quad -\quad \cancel{97.96.95.94}\\ \quad$

$=98.99.100.101$

Hence, $y=\frac{1}{4}*98.99.100.101=\boxed{\color{#3D99F6}{24497550}}$

$\begin{aligned} S & = 1\dot{} 2 \dot{} 3 + 2 \dot{} 3 \dot{} 4 + 3 \dot{} 5 \dot{} 6 + ... + 98 \dot{} 99 \dot{} 100 \\ & = \sum_{n=1}^{98} n(n+1)(n+2) \\ & = \sum_{n=1}^{98} (n^3 + 3n^2 + 2n) \\ & = \left( \frac{98(99)}{2} \right)^2 + 3 \times \frac{98(99)(197)}{6} + 2 \times \frac{98(99)}{2} \\ & = 49(99)\left( 49(99) + 197 + 2\right) \\ & = \boxed{24497550} \end{aligned}$

Note that $(n-1)\cdot n\cdot (n+1) = n (n^2-1)$ . Therefore, $0\cdot 1\cdot 2 + 1\cdot 2\cdot 3 + 2\cdot 3 \cdot 4 + \cdots + 98\cdot99\cdot 100\\ = 1\cdot (1^2-1) + 2\cdot (2^2-1) + 3\cdot (3^2-1) + \cdots + 99\cdot (99^2-1) \\ = (1^3 + 2^3 + \cdots + 99^3) - (1 + 2 + \cdots 99) \\ = (1 + 2 + \cdots + 99)^2 - (1 + 2 + \cdots 99) \\ = 4950^2 - 4950 = 4949\cdot 4950 = \boxed{24\:497\:550}.$

Notes: I added the term $0 \cdot 1 \cdot 2$ for convenience. It does not affect the sum.

Also, I assumed as a known fact that $\sum n^3 = \left(\sum n\right)^2$ .

$\large {T_n=n(n+1)(n+2)}$

$\large {T_n=\frac{1}{4} n(n+1)(n+2)[(n+3)-(n-1)]}$

$\frac {1}{4}$ [ $\large {n(n+1)(n+2)(n+3)}$ - $\large {(n-1)(n)(n+1)(n+2)}$ ]

Now, Solving this telescopic series,we get

= $\boxed{\frac{99\times 98\times 100\times 101}{4}}$

= $Ans.$ $\boxed{24497550}$

I used the formula: x⁴/4 + 3x³/2 + 11x²/4 +3x/2 . x=98 , Answer is 24497550 . But they have easier solution on that :)

It equals $6(\binom{3}{3}+\binom{4}3+\binom53+...+\binom{100}3)=6\binom{101}4=\frac{101\times100\times99\times98}4$

We have: $(n-1)(n)(n+1)= n^3-n$
Now using the above formula the equation becomes:
$2^3-2 + 3^3-3............99^3-99$
$=(1+2+3+4.......+99)^2-1-4949$
$=4950^2 - 4950$ $=\boxed{24497550}$

I noted a pattern that $1^3+2^3......+n^3= (1+2+3....+n)^2$ and used the same above.This may already be a theorem or something that i have never heard of.