I am looking through the mark scheme of a past A level paper and I cannot work out how they have simplified an expression:
$$\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{1}{3}(k+1)(k+2)(k+3)$$
Can anyone walk me through it?
Thanks.
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1$\begingroup$ Consider getting a common denominator then factoring out like terms. $\endgroup$ – kmeis Jan 26 '17 at 18:05
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We have: $$\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$ Then: $$\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3}$$ We can now combine the two fractions: $$\frac{\color{blue}{k}\color{#cc0000}{(k+1)(k+2)}+\color{blue}{3}\color{#cc0000}{(k+1)(k+2)}}{3}$$ Note that $(k+1)(k+2)$ are alike on both. Hence, we can factorise: $$\frac{\color{blue}{(k+3)}\color{#cc0000}{(k+1)(k+2)}}{3}$$ Which is the answer given.
answered Jan 26 '17 at 18:06
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Take $(k+1)(k+2)$ common, we have -
$(k+1)(k+2) \left[\frac k3 + 1\right]$
= $(k+1)(k+2) \left[\frac{k+3}{3}\right]$
= $\frac13(k+1)(k+2)(k+3)$
answered Jan 26 '17 at 18:17
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$\begin{array}{rcl}\dfrac13k(k+1)(k+2)+(k+1)(k+2) &=& (k+1)(k+2)\left(\dfrac13k+1\right) \\ &=&(k+1)(k+2)\left(\dfrac13k+\dfrac13\cdot3\right) \\&=& \dfrac13(k+1)(k+2)(k+3) \end{array}$
