This expression can be rewritten as

$3*2 + 4*3 + 5*4 + 6*5 + ... + 50*49 =$

$\displaystyle\sum_{n=2}^{49} n(n + 1) = -2 + \displaystyle\sum_{n=1}^{49} n(n + 1) =$

$-2 + \displaystyle\sum_{n=1}^{49} n^{2} + \sum_{n=1}^{49} n = -2 + \dfrac{49(49 + 1)(2*49 + 1)}{6} + \dfrac{49(49 + 1)}{2} =$

$-2 + (49)(25)(33) + (49)(25) = (49)(25)(34) - 2 = \boxed{41648}.$

We can write this as

$2\times3+3\times4+4\times5+\cdots+49\times50$

$=\left(\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{50}{2}\right)\cdot2$

$=2\left(\binom{51}{3}-\binom{2}{2}\right)=41648$

Where we used the hockey stick identity from the transition from the second to third line.

• This sum can be rewritten as $2\times3 + 3\times4 + 4\times5 + ... + 49\times50 = 6 + 12 + 20 + 30 .....$ .
• The sequence (6, 12, 20, 30, 42, ...) is a second-degree arithmetical progression; it means the sum of the n first terms can be written as a third-degree polinomy.
• We have: $A\times n^{3} + B\times n^{2} + C\times n + D = S_n$ .
• We will use this expression a couple times now to create a system of equations.

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• For n = 1 , $S_1 = 6$ ; for n = 2 , $S_2 = (6 + 12) = 18$ ; for n = 3 , $S_3 = (6 + 12 + 20) = 38$ ; and for n = 4 , $S_4 = (6 + 12 + 20 + 30) = 68$ ; so further, we have:

1. I , for n = 1, $(A + B + C + D) = 6$ .
2. II , for n = 2, $(8A + 4B + 2C + D) = 18$ .
3. III , for n = 3, $(27A + 9B + 3C + D) = 38$ .

4. IV , for n = 4, $(64A + 16B + 4C + D) = 68$ .

   Subtracting I from II, II from III and III from IV, we afford:

5. V , $(7A + 3B + C) = 12$ .
6. VI , $(19A + 5B + C) = 20$ .

7. VII , $(37A + 7B + C) = 30$ .

   Subtracting V from VI and VI from VII, we have:

8. VIII , $(12A + 2B) = 8$ .
9. IX , $(18A + 2B) = 10$ .

• FINALLY, subtracting VIII from IX, we state that $6A = 2 => A = \frac{1}{3}$ .

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• Replacing $A = \frac{1}{3}$ in VIII, we state B = 2 .
• Doing the same process all over, we conclude: $A = \frac{1}{3}, B = \frac{6}{3}, C = \frac{11}{3}$ and D = 0 .
• After all this process, we built our expression as $S_n = \frac{n^{3} + 6n^{2} + 11n}{3}$ .
• Now, concluding, we use n = 48 and state that this whole sum equals 41648 .

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• Guys, I know this seems like a pretty pretty complex solution, but it is only hard to explain, as it goes so quickly with hand and pen. I tried to write the whole and complete detailed solution, so everything get explained; however, if you have any doubts, questions or comments, you can write it below and I swear i'll do my best to help you all.

The general term is

(n + 2)!/n! = (n + 2)(n +1) = n^2 + 3 n + 2 ,

where n = 1 , 2 , 3, ......... , 48

Sum of (n^2) = n(n + 1)(2 n + 1)/6 = (48)(49)(97)/6 = 38024

Sum of (3 n) = (3) n(n + 1) = (3)(48)(49) = 3528

Sum of (2) =2 n = (2)(48) = 96

The total sum = 38024 + 3528 + 96 = 41648

include <iostream>

using namespace std; int main() { int a=0; int z=1; int y=1; int x=1; for(x=1;x<=48;x++) { while(x==1) { y=(x+1) (x+2); cout << y << "\t" << endl; z=a+y; break;} while(x>=2 && x<=48) { y=(x+1) (x+2); if (x==2) z=a+y; else { z+=y;} cout << y << "\t" << z << endl; break; } a=y; } cin.ignore(); cin.ignore(); return 0; }

2 3+3 4+•••+49 50=1/3((-1 2 3+2 3 4)+(-2 3 4+3 4 5)+•••+(-48 49 50+49 50 51))=1/3(-1 2 3+49 50*51)=41648

This sum can be thought of as Sigma of (r+2)!/r! =(r+1)*(r+2) This is a simple telescopic setup. Upon solving, we get 1/3(49 * 50 * 51-1 * 2 * 3) =41648