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# Calculus/Precalculus

Question
How would you solve this using Mathematical Induction?

1 + 3 + 6 . . . + n(n+1)/(2) = n(n+1)(n+2)/(6)

For n = 1, we have 1 = 1*2*3/6 = 1, so it is true for n = 1.

Assume it is true for n = k.  That is, 1 + 3 + 6 . . . + k(k+1)/(2) = k(k+1)(k+2)/6.

For n = k + 1, we have That is, 1 + 3 + 6 . . . + k(k+1)/2 + (k+1)(k+2)/2 on the left side.
It is known that the first k terms have the sum k(k+1)(k+2)/6.
This means the sum is really k(k+1)(k+2)/6 + (k+1)(k+2)/2.

We can see k(k+1) = k²+k, and (k²+k)(k+2) = k³ + 3k² + 2k and (k+1)(k+2) = k²+3k+2.
This means we have (k³ + 3k² + 2k)/6 + (k² + 3k + 2)/2.

Since the latter only has a 2 in the denominator and needs a 6, multiply by 3/3.
This gives (k³ + 3k² + 2k + 3k² + 9k + 6)/6.  Combining like terms gives
(k³ + 6k² + 11k + 6)/6.  Factoring out k+1 on top gives (k+1)(k²+5k+6)/6.
Factoring out k+2 on top gives (k+1)(k+2)(k+3)/6, and that is what we get when we replace
n with k+1 in  n(n+1)(n+2)/6.

Since I have shown it to be true at n=1, assumed it was true at n=k,
and proved it was true at n=k+1, this is true for all integers n >= 1.

Calculus

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#### Scotto

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