# Show # 1.2.3 + 2.3.4+ 3.4.5 + ... + n(n+1)(n+2) = 1/4n(n+1)(n+2)(n+3)#?

##### 2 Answers

The sum is:

# S_n = 1.2.3 + 2.3.4+ 3.4.5 + ... + n(n+1)(n+2) #

# " " = sum_(r=1)^n r(r+1)(r+2) #

# " " = sum_(r=1)^n r(r^2+3r+2) #

# " " = sum_(r=1)^n (r^3+3r^2+2r) #

# " " = sum_(r=1)^n r^3 + 3sum_(r=1)^nr^2+2sum_(r=1)^nr #

We now use some standard formula for summation:

# sum_(r=1)^n r \ = 1/2n(n+1) #

# sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #

# sum_(r=1)^n r^3 = 1/4n^2(n+1)^2 #

Which gives us:

# S_n = 1/4n^2(n+1)^2 + 3*1/6n(n+1)(2n+1) + 2*1/2n(n+1) #

# " " = 1/4n^2(n+1)^2 + 1/2n(n+1)(2n+1) + n(n+1) #

# " " = 1/4n(n+1){n(n+1) +2(2n+1) + 4} # >

# " " = 1/4n(n+1)(n^2+n +4n+2 + 4) #

# " " = 1/4n(n+1)(n^2+5n+6) #

# " " = 1/4n(n+1)(n+2)(n+3) \ \ # QED

We can also prove the given result using Mathematical Induction.

Let:

# S_n = 1.2.3 + 2.3.4+ 3.4.5 + ... + n(n+1)(n+2) #

# " " = sum_(r=1)^n r(r+1)(r+2) #

We want to prove that:

# S_n = 1/4n(n+1)(n+2)(n+3) \ \ \ AA n in NN #

Let use consider the case

# LHS = S_1#

# " " = 1.2.3 #

# " " = 6 #

And:

# RHS = 1/4(1)(1+1)(1+2)(1+3) #

# " " = (1.2.3.4)/4 #

# " " = 24/4 #

# " " = 6 #

So the given result is **true** when

Now, Let us **assume** that the given result is true when

# S_k = sum_(r=1)^k r(r+1)(r+2) = 1/4k(k+1)(k+2)(k+3) #

Adding the next term to the series gives us:

# S_(k+1) = S_k + (k+1)(k+2)(k+3) #

# " " = 1/4k(k+1)(k+2)(k+3) + (k+1)(k+2)(k+3) # (by assumption)

# " " = (k(k+1)(k+2)(k+3))/4 + (4(k+1)(k+2)(k+3))/4 #

# " " = 1/4(k+1)(k+2)(k+3)(k+4)#

# " " = 1/4 \ (k+1) \ {(k+1)+1} \ {(k+1)+2} \ {(k+1)+3}#

Which is the given result with

So, we have shown that if the given result is true for

Hence, by the process of mathematical induction the given result is true for