proving summation using induction

Question

we need to prove that for ever $n∈N$ the following equality is right $$\frac{3\sum_{i=1}^n i^5}{{\sum_{i=1}^n i^3}} = 2n^2+2n-1$$ so first of i checked for n=1 and we get $$\frac{3\sum_{i=1}^1 i^5}{{\sum_{i=1}^1 i^3}} = 2*1^2+2*1-1$$ which is $3=3$

now I assume for $n=k$

$$\frac{3\sum_{i=1}^k i^5}{{\sum_{i=1}^k i^3}} = 2k^2+2k-1$$ and next is to solve for $n=k+1$ $$\frac{3\sum_{i=1}^{k+1} (i^5)}{{\sum_{i=1}^{k+1} i^3}} = 2(k+1)^2+2(k+1)-1$$

$$\frac{3\sum_{i=1}^{k+1} (i^5)}{{\sum_{i=1}^{k+1} i^3}} = 2k^2+6k+3$$

now I tried to separate the $\sum_{i=1}^{k+1}$

$$\frac{3\sum_{i=1}^{k} (i^5)+{3\sum_{i=k+1}^{k+1} (i^5)}}{{\sum_{i=1}^{k} i^3}+{\sum_{i=k+1}^{k+1} (i^3)}} = 2k^2+6k+3$$

now since $$\frac{3\sum_{i=1}^k i^5}{{\sum_{i=1}^k i^3}} = 2k^2+2k-1$$ i tried putting it instead of what i have in the fraction and i got stuck

$$\frac {2k^2+2k-1+{3\sum_{i=k+1}^{k+1} (i^5)}}{{\sum_{i=k+1}^{k+1} (i^3)}} = 2k^2+6k+3$$

sorry for my English mistakes hope it is understandable , appreciate any help and tips!

you could try proving formulas for the sum of $5$th and $3$rd powers separately and then take the ratio — J. W. Tanner, Nov 13 '20 at 18:37
@J.W.Tanner thank you for the tip but i did not understand what you mean by proving formulas for the 5th and 3rd powers? what is there to prove in a sum like this ? thank you much appreciated — Adamrk, Nov 13 '20 at 18:51

score 1 · Accepted Answer · answered Nov 13 '20 at 18:59

1

Let us denote $\sum_{i=1}^k i^3$ by $S_3$ and $\sum_{i=1}^k i^5$ by $S_5$

Given $$\frac{3S_5}{S_3} = 2k^2+2k -1$$

you are required to prove (RTP) that

$$\frac{3S_5 + (k+1)^5}{S_3 + (k+1^3)} = 2k^2+6k +3$$

In the numerator of the above expression, put $$3S_5 = S_3(2k^2+2k -1)$$ and solve for $S_3$

You'll find that $$S_3 = \frac{k^2(k+1)^2}{4}$$ which is always true.

That completes the proof.

answered Nov 13 '20 at 18:59

PTDS

3,464

amazing thank you !! – Adamrk Nov 13 '20 at 19:03

proving summation using induction

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