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Mathematics 20 Online
OpenStudy (anonymous):

what is math

OpenStudy (anonymous):

can someone help me with this?

OpenStudy (anonymous):

Im like really lost

OpenStudy (unimatix):

No one can help you understand what math is. Math is zen. It is the yen to the yang. It is normal to feel lost. But continue on in this dark path in the forest and one day, perhaps you will emerge young neophyte, as but a wizard. A wizard trained in the dark magical arts of wizardry that is math.

hero (hero):

Math is an alien language that very few really and fully understand.

OpenStudy (venomblast):

It's Life

OpenStudy (irishboy123):

it's a language with no cuss words

Nnesha (nnesha):

`1st)` substitute n for 1 to check is the L.H.S = R.H.S ? if the statement is true then next step is `2nd)` assume it is true for n=k (substitute n for k) this step is called " induction assumption' `3rd)` substitute n for k+1 and we want to show the statement is true for n= k+1 based on the 2nd step assumption here is an example \[\large\rm 3+8+13+18+....(5n-2)=\frac{ n(5n+1) }{ 2 }\] show `n=1` substitute n for 1 \[\large\rm (5(\color{ReD}{1})-2)=\frac{ \color{reD}{1}(5(\color{ReD}{1})+1) }{ 2 }\]\[3=\frac{ 6 }{ 2 } ~~~~~~~3=3\] both sides are equal so `n=1` is true 2nd) step assume n =k \[\large\rm \color{orange}{ 3+8+13+18+....(5\color{ReD}{k}-2)=\frac{ \color{Red}{k}(5\color{Red}{k}+1) }{ 2 }}\] assume it's true now substitute n for k+1 \[\rm 3+8+13+18+....(5k-2)\color{blue}{+}(5(\color{ReD}{k+1})-2)=\frac{ \color{Red}{(k+1)}(5(\color{Red}{k+1})+1) }{ 2 }\] distributive property \[\rm 3+8+13+18+....(5k-2)\color{blue}{+}(5k+5-2)=\frac{ \color{Red}{(k+1)}(5k+5+1) }{ 2 }\] hmm simplify \[\rm \color{orange}{ 3+8+13+18+....(5k-2)}\color{blue}{+}(5k+3)=\frac{ \color{black}{(k+1)}(5k+6) }{ 2 }\] `3+8+13+18+....(5k-2) = k(5k+1)/2 ` so plugin \[\rm \color{orange}{ \frac{ \color{Red}{k}(5\color{Red}{k}+1) }{ 2 }}\color{blue}{+}(5k+3)=\frac{ \color{black}{(k+1)}(5k+6) }{ 2 }\]

Nnesha (nnesha):

\[\rm \color{orange}{ \frac{ \color{Red}{k}(5\color{Red}{k}+1) }{ 2 }}\color{blue}{+}(5k+3)=\frac{ \color{black}{(k+1)}(5k+6) }{ 2 }\] if we solve both sides you will see L.H.S= R.H.S

Nnesha (nnesha):

let me know if you don't understand any step or if you have a question abt that example. goooood reviewed \o^_^o/

Nnesha (nnesha):

\(\huge\color{green}{\rm {Math~is~FUN!!:)}}\)\(\Huge \color{orange}{\star^{ \star^{\star:)}}}\Huge \color{purple}{\star^{ \star^{\star:)}}}\)\(\rm\color{green}{o^\wedge\_^\wedge o}\)

OpenStudy (anonymous):

@Nnesha Thank you so much! I understand it a lot better now :)

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