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
Apr 1, 2017

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

Apr 1, 2017

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 #n=1#. When #n=1# the given result gives:

# 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 #n=1#

Now, Let us assume that the given result is true when #n=k#, for some #k in NN#, in which case for this particular value of #k# we have:

# 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 #n=k+1#

So, we have shown that if the given result is true for #n=k#, then it is also true for #n=k+1#. But we initially showed that the given result was true for #n=1# and so it must also be true for #n=2, n=3, n=4, ... # and so on.

Hence, by the process of mathematical induction the given result is true for #n in NN# QED