Question Number 59133 by Tawa1 last updated on 05/May/19
$$\mathrm{The}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression}\:\:\:\left(\mathrm{x}\:+\:\mathrm{a}\right)^{\mathrm{100}} \:+\:\left(\mathrm{x}\:−\:\mathrm{a}\right)^{\mathrm{100}} \\ $$$$\mathrm{after}\:\mathrm{simplification}\:\mathrm{is}\:\:? \\ $$
Answered by $@ty@m last updated on 05/May/19
$${In}\:\left({x}−{a}\right)^{{n}} , \\ $$$${No}.\:{of}\:{negative}\:{terms}=\begin{cases}{\frac{{n}+\mathrm{1}}{\mathrm{2}}\:{if}\:{n}\:{is}\:{odd}.}\\{\frac{{n}}{\mathrm{2}}\:{if}\:{n}\:{is}\:{even}.}\end{cases} \\ $$$${All}\:{negative}\:{terms}\:\:{of}\:\left({x}−{a}\right)^{{n}} \:{cancel} \\ $$$${with}\:{like}\:{positive}\:{terms}\:{of}\:\left({x}+{a}\right)^{{n}} . \\ $$$$\therefore\:\mathrm{The}\:\:\mathrm{number}\:\mathrm{of}\:{cancelled}\:\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression}\:\:\:\left(\mathrm{x}\:+\:\mathrm{a}\right)^{{n}} \:+\:\left(\mathrm{x}\:−\:\mathrm{a}\right)^{{n}} \:{is} \\ $$$$\begin{cases}{\mathrm{2}\left({n}+\mathrm{1}\right)\:−\mathrm{2}×\frac{{n}+\mathrm{1}}{\mathrm{2}}\:{if}\:{n}\:{is}\:{odd}}\\{\mathrm{2}\left({n}+\mathrm{1}\right)−\mathrm{2}×\frac{{n}}{\mathrm{2}}\:{if}\:{n}\:{is}\:{even}.}\end{cases} \\ $$$$=\begin{cases}{{n}+\mathrm{1}\:{if}\:{n}\:{is}\:{odd}}\\{{n}+\mathrm{2}\:{if}\:{n}\:{is}\:{even}.}\end{cases} \\ $$$$\therefore\:\mathrm{The}\:\:{total}\:\mathrm{number}\:\mathrm{of}\:\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:{v}\mathrm{expression}\:\:\:\left(\mathrm{x}\:+\:\mathrm{a}\right)^{{n}} \:+\:\left(\mathrm{x}\:−\:\mathrm{a}\right)^{{n}} \\ $$$${after}\:{addition}\:{of}\:{like}\:{terms} \\ $$$$=\begin{cases}{\frac{{n}+\mathrm{1}}{\mathrm{2}}\:{if}\:{n}\:{is}\:{odd}}\\{\frac{{n}+\mathrm{2}}{\mathrm{2}}\:{if}\:{n}\:{is}\:{even}.}\end{cases} \\ $$$${When}\:{n}=\mathrm{100} \\ $$$${Req}.{no}.\:{of}\:{terms}=\frac{\mathrm{102}}{\mathrm{2}}=\mathrm{51}. \\ $$
Answered by mr W last updated on 05/May/19
![(x+a)^(100) =Σ_(k=0) ^(100) C_k ^(100) a^(100−k) x^k (x−a)^(100) =Σ_(k=0) ^(100) C_k ^(100) (−1)^(100−k) a^(100−k) x^k (x+a)^(100) +(x−a)^(100) =Σ_(k=0) ^(100) C_k ^(100) [1+(−1)^(100−k) ]a^(100−k) x^k 1+(−1)^(100−k) =2 for k=0,2,...,100 ⇒51 terms 1+(−1)^(100−k) =0 for k=1,3,...,99 ⇒50 terms ⇒(x+a)^(100) +(x−a)^(100) has 51 terms.](https://www.tinkutara.com/question/Q59151.png)
$$\left({x}+{a}\right)^{\mathrm{100}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{100}} {\sum}}{C}_{{k}} ^{\mathrm{100}} {a}^{\mathrm{100}−{k}} {x}^{{k}} \\ $$$$\left({x}−{a}\right)^{\mathrm{100}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{100}} {\sum}}{C}_{{k}} ^{\mathrm{100}} \left(−\mathrm{1}\right)^{\mathrm{100}−{k}} {a}^{\mathrm{100}−{k}} {x}^{{k}} \\ $$$$\left({x}+{a}\right)^{\mathrm{100}} +\left({x}−{a}\right)^{\mathrm{100}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{100}} {\sum}}{C}_{{k}} ^{\mathrm{100}} \left[\mathrm{1}+\left(−\mathrm{1}\right)^{\mathrm{100}−{k}} \right]{a}^{\mathrm{100}−{k}} {x}^{{k}} \\ $$$$\mathrm{1}+\left(−\mathrm{1}\right)^{\mathrm{100}−{k}} =\mathrm{2}\:{for}\:{k}=\mathrm{0},\mathrm{2},…,\mathrm{100}\:\Rightarrow\mathrm{51}\:{terms} \\ $$$$\mathrm{1}+\left(−\mathrm{1}\right)^{\mathrm{100}−{k}} =\mathrm{0}\:{for}\:{k}=\mathrm{1},\mathrm{3},…,\mathrm{99}\:\Rightarrow\mathrm{50}\:{terms} \\ $$$$ \\ $$$$\Rightarrow\left({x}+{a}\right)^{\mathrm{100}} +\left({x}−{a}\right)^{\mathrm{100}} \:{has}\:\mathrm{51}\:{terms}. \\ $$
Commented by $@ty@m last updated on 05/May/19
$${Thanks}\:{for}\:{correction}…. \\ $$
Commented by Tawa1 last updated on 05/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$