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Prove by mathematical induction that $$1\cdot2+3\cdot4+\cdots+(2n−1)\cdot2n=n(n+1)(4n−1)/3$$ for all $n\in\mathbb{N}$.

Let claim (n) be $$(1\cdot 2) + (3\cdot4) +\cdots+(2n-1)\cdot2n = n(n+1)(4n-1)/3$$

claim(1): $$(2(1)-1)\cdot(2)(1)=1(1+1)(4(1)-1)/3\\ 2=2$$

Hence claim (k): $$(1\cdot2)+(3\cdot4)+....+(2k-1)\cdot2k=k(k+1)(4k-1)/3$$

claim $(k+1)$: RHS$= (k+1)(k+2)(4k+3)/3$

claim $(k+1)$ LHS: $$= (1\cdot2)+(3\cdot4)+(2k-1)\cdot2k +(2(k+1))\cdot(2(k+1))\\ = (1\cdot2)+(3\cdot4)+(2k-1)\cdot2k +(2k+2)\cdot(2k+2)$$

No matter how i try i cannot prove the LHS = RHS!!! Can anyone help to show me how i prove the LHS = RHS by using mathematical induction

Thomas Andrews
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Kelvin Ckl
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3 Answers3

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for $n=1$ your Statement is true. We suppose that for $n=k$ holds $$1\cdot 2+3\cdot4 +...+(2k-1)2k=\frac{k(k+1)(4k-1)}{3}$$ now we have to prove that$$1\cdot 2+3\cdot 4+...+(2k-1)2k+(2k+1)2(k+1)=\frac{(k+1)(k+2)(4k+3)}{3}$$ we have $$\frac{k(k+1)(4k-1)}{3}+(2k+1)2(k+1)=\frac{(k+1)(4k^2+11k+6)}{3}=\frac{(k+1)(k+2)(4k+3)}{3}$$

Dr. Sonnhard Graubner
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For $n=1$ it's true and since after assumption of the induction we see that $$1\cdot2+...+(2n-1)\cdot2n+(2n+1)(2n+2)=$$ $$=\frac{n(n+1)(4n-3)}{3}+(2n+1)(2n+2)=\frac{(n+1)(n+2)(4n+3)}{3},$$ we are done!

The last step it's $$n(4n-3)+6(2n+1)=(n+2)(4n+3)$$ or $$4n^2-3n+12n+6=4n^2+9n+6,$$ which is obvious.

Michael Rozenberg
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We want to prove that

$$\sum\limits_{j=1}^{n+1} 2j(2j-1)=\frac {(n+1)} 3 (n+2)(4n+3)$$

By induction hypothesis we have,

$$\sum\limits_{j=1}^{n} 2j(2j-1)= \frac {n(n+1)(4n-1)} 3$$

$$\sum\limits_{j=1}^{n+1} 2j(2j-1)= \frac {n(n+1)(4n-1)} 3 + 2(2n+1)(n+1)$$

$$\sum\limits_{j=1}^{n+1} 2j(2j-1)=\frac {(n+1)} 3 ( {n(4n-1)} + 6(2n+1))$$

$$\sum\limits_{j=1}^{n+1} 2j(2j-1)=\frac {(n+1)} 3 P(n)$$

Where $P(n)=4n^2+11n+6$. We see that P(-2)=0 then $P(n)=(n+2)(4n+3)$

Therefore,

$$\sum\limits_{j=1}^{n+1} 2j(2j-1)=\frac {(n+1)} 3 (n+2)(4n+3)$$

user577215664
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