Menu Close

Question-173776




Question Number 173776 by AgniMath last updated on 18/Jul/22
Answered by thfchristopher last updated on 18/Jul/22
When n=1,  R.H.S.=((1×2)/2)  =1  =L.H.S.  Assume n=k is true,  R.H.S.=((k(k+1))/2)  When n=k+1,  R.H.S.=((k(k+1))/2)+k+1  =((k^2 +k+2k+2)/2)  =((k^2 +3k+2)/2)  =(((k+1)(k+2))/2)  ∴ n=k+1 is true.  Inductively, n is true for all natural values.
$$\mathrm{When}\:{n}=\mathrm{1}, \\ $$$$\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{1}×\mathrm{2}}{\mathrm{2}} \\ $$$$=\mathrm{1} \\ $$$$=\mathrm{L}.\mathrm{H}.\mathrm{S}. \\ $$$$\mathrm{Assume}\:{n}={k}\:\mathrm{is}\:\mathrm{true}, \\ $$$$\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{When}\:{n}={k}+\mathrm{1}, \\ $$$$\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}+{k}+\mathrm{1} \\ $$$$=\frac{{k}^{\mathrm{2}} +{k}+\mathrm{2}{k}+\mathrm{2}}{\mathrm{2}} \\ $$$$=\frac{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{2}}{\mathrm{2}} \\ $$$$=\frac{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\therefore\:{n}={k}+\mathrm{1}\:\mathrm{is}\:\mathrm{true}. \\ $$$$\mathrm{Inductively},\:{n}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{all}\:\mathrm{natural}\:\mathrm{values}. \\ $$
Answered by MJS_new last updated on 18/Jul/22
1. n is even     1       2          3       4 ...     n    n−1 n−2 n−3 ...  ============  n+1 n+1  n+1 n+1 ... (n/2) times  ⇒ sum is (n/2)(n+1)  2. n is uneven     1       2          3       4 ...     n    n−1 n−2 n−3 ...  ============  n+1 n+1  n+1 n+1 ... ((n−1)/2) times and ((n+1)/2) is “left alone”  ⇒ sum is ((n−1)/2)(n+1)+(1/2)(n+1)=(n/2)(n+1)
$$\mathrm{1}.\:{n}\:\mathrm{is}\:\mathrm{even} \\ $$$$\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\mathrm{4}\:… \\ $$$$\:\:\:{n}\:\:\:\:{n}−\mathrm{1}\:{n}−\mathrm{2}\:{n}−\mathrm{3}\:… \\ $$$$============ \\ $$$${n}+\mathrm{1}\:{n}+\mathrm{1}\:\:{n}+\mathrm{1}\:{n}+\mathrm{1}\:…\:\frac{{n}}{\mathrm{2}}\:\mathrm{times} \\ $$$$\Rightarrow\:\mathrm{sum}\:\mathrm{is}\:\frac{{n}}{\mathrm{2}}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{2}.\:{n}\:\mathrm{is}\:\mathrm{uneven} \\ $$$$\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\mathrm{4}\:… \\ $$$$\:\:\:{n}\:\:\:\:{n}−\mathrm{1}\:{n}−\mathrm{2}\:{n}−\mathrm{3}\:… \\ $$$$============ \\ $$$${n}+\mathrm{1}\:{n}+\mathrm{1}\:\:{n}+\mathrm{1}\:{n}+\mathrm{1}\:…\:\frac{{n}−\mathrm{1}}{\mathrm{2}}\:\mathrm{times}\:\mathrm{and}\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\mathrm{is}\:“\mathrm{left}\:\mathrm{alone}'' \\ $$$$\Rightarrow\:\mathrm{sum}\:\mathrm{is}\:\frac{{n}−\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right)=\frac{{n}}{\mathrm{2}}\left({n}+\mathrm{1}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *