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Question Number 59133 by Tawa1 last updated on 05/May/19

The total number of terms in the expression   (x + a)^(100)  + (x − a)^(100)   after simplification is  ?

Thetotalnumberoftermsintheexpression(x+a)100+(xa)100aftersimplificationis?

Answered by $@ty@m last updated on 05/May/19

In (x−a)^n ,  No. of negative terms= { ((((n+1)/2) if n is odd.)),(((n/2) if n is even.)) :}  All negative terms  of (x−a)^n  cancel  with like positive terms of (x+a)^n .  ∴ The  number of cancelled  terms in the expression   (x + a)^n  + (x − a)^n  is   { ((2(n+1) −2×((n+1)/2) if n is odd)),((2(n+1)−2×(n/2) if n is even.)) :}  = { ((n+1 if n is odd)),((n+2 if n is even.)) :}  ∴ The  total number of  terms in the vexpression   (x + a)^n  + (x − a)^n   after addition of like terms  = { ((((n+1)/2) if n is odd)),((((n+2)/2) if n is even.)) :}  When n=100  Req.no. of terms=((102)/2)=51.

In(xa)n,No.ofnegativeterms={n+12ifnisodd.n2ifniseven.Allnegativetermsof(xa)ncancelwithlikepositivetermsof(x+a)n.Thenumberofcancelledtermsintheexpression(x+a)n+(xa)nis{2(n+1)2×n+12ifnisodd2(n+1)2×n2ifniseven.={n+1ifnisoddn+2ifniseven.Thetotalnumberoftermsinthevexpression(x+a)n+(xa)nafteradditionofliketerms={n+12ifnisoddn+22ifniseven.Whenn=100Req.no.ofterms=1022=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.

(x+a)100=100k=0Ck100a100kxk(xa)100=100k=0Ck100(1)100ka100kxk(x+a)100+(xa)100=100k=0Ck100[1+(1)100k]a100kxk1+(1)100k=2fork=0,2,...,10051terms1+(1)100k=0fork=1,3,...,9950terms(x+a)100+(xa)100has51terms.

Commented by $@ty@m last updated on 05/May/19

Thanks for correction....

Thanksforcorrection....

Commented by Tawa1 last updated on 05/May/19

God bless you sir

Godblessyousir

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