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Question Number 1074 by Yugi last updated on 05/Jun/15

Prove by induction the following result where N is a positive even integer.                                                               S_1 ^2 +S_2 ^2 =2^N   where    S_1 =Σ_(r=1) ^((N/2)+1) (−1)^(r−1)  ((N),((2(r−1))) )    and      S_2 =Σ_(r=1) ^(N/2) (−1)^(r+1)  ((N),((2r−1)) )     .

$${Prove}\:{by}\:{induction}\:{the}\:{following}\:{result}\:{where}\:{N}\:{is}\:{a}\:{positive}\:{even}\:{integer}.\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{S}_{\mathrm{1}} ^{\mathrm{2}} +{S}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{2}^{{N}} \\ $$$${where}\:\:\:\:{S}_{\mathrm{1}} =\underset{{r}=\mathrm{1}} {\overset{\frac{{N}}{\mathrm{2}}+\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} \begin{pmatrix}{{N}}\\{\mathrm{2}\left({r}−\mathrm{1}\right)}\end{pmatrix}\:\:\:\:{and}\:\:\:\:\:\:{S}_{\mathrm{2}} =\underset{{r}=\mathrm{1}} {\overset{{N}/\mathrm{2}} {\sum}}\left(−\mathrm{1}\right)^{{r}+\mathrm{1}} \begin{pmatrix}{{N}}\\{\mathrm{2}{r}−\mathrm{1}}\end{pmatrix}\:\:\:\:\:. \\ $$

Commented by prakash jain last updated on 07/Jun/15

^n C_r +^n C_(r−1) =^(n+1) C_r   ^(n+2) C_r =^(n+1) C_(r−1) +^(n+1) C_r =^n C_(r−2) +2^n C_(r−1) +^n C_r

$$\:^{{n}} {C}_{{r}} +\:^{{n}} {C}_{{r}−\mathrm{1}} =^{{n}+\mathrm{1}} {C}_{{r}} \\ $$$$\:^{{n}+\mathrm{2}} {C}_{{r}} =\:^{{n}+\mathrm{1}} {C}_{{r}−\mathrm{1}} +\:^{{n}+\mathrm{1}} {C}_{{r}} =\:^{{n}} {C}_{{r}−\mathrm{2}} +\mathrm{2}\:^{{n}} {C}_{{r}−\mathrm{1}} +\:^{{n}} {C}_{\mathrm{r}} \\ $$

Commented by prakash jain last updated on 07/Jun/15

Use the above relation to write S_1 ^(N+2) in terms  of S_1 ^N and S_2 ^N .

$$\mathrm{Use}\:\mathrm{the}\:\mathrm{above}\:\mathrm{relation}\:\mathrm{to}\:\mathrm{write}\:\overset{{N}+\mathrm{2}} {{S}}_{\mathrm{1}} \mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\overset{{N}} {{S}}_{\mathrm{1}} \mathrm{and}\:\overset{{N}} {{S}}_{\mathrm{2}} . \\ $$

Commented by 112358 last updated on 07/Jun/15

Thanks.

$${Thanks}.\: \\ $$

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