The-total-number-of-terms-in-the-expression-x-a-100-x-a-100-after-simplification-is-

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 ?

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

I n {(x - a)}^{n},

N o . o f n e g a t i v e t e r m s = {\begin{cases} \frac{n + 1}{2} i f n i s o d d . \\ \frac{n}{2} i f n i s e v e n . \end{cases}

A l l n e g a t i v e t e r m s o f {(x - a)}^{n} c a n c e l

w i t h l i k e p o s i t i v e t e r m s o f {(x + a)}^{n} .

∴ The number of c a n c e l l e d terms in the expression {(x + a)}^{n} + {(x - a)}^{n} i s

{\begin{cases} 2 (n + 1) - 2 \times \frac{n + 1}{2} i f n i s o d d \\ 2 (n + 1) - 2 \times \frac{n}{2} i f n i s e v e n . \end{cases}

= {\begin{cases} n + 1 i f n i s o d d \\ n + 2 i f n i s e v e n . \end{cases}

∴ The t o t a l number of terms in the v expression {(x + a)}^{n} + {(x - a)}^{n}

a f t e r a d d i t i o n o f l i k e t e r m s

= {\begin{cases} \frac{n + 1}{2} i f n i s o d d \\ \frac{n + 2}{2} i f n i s e v e n . \end{cases}

W h e n n = 100

R e q . n o . o f t e r m s = \frac{102}{2} = 51 .

Answered by mr W last updated on 05/May/19

{(x + a)}^{100} = \underset{k = 0}{\sum^{100}} C_{k}^{100} a^{100 - k} x^{k}

{(x - a)}^{100} = \underset{k = 0}{\sum^{100}} C_{k}^{100} {(- 1)}^{100 - k} a^{100 - k} x^{k}

{(x + a)}^{100} + {(x - a)}^{100} = \underset{k = 0}{\sum^{100}} C_{k}^{100} [1 + {(- 1)}^{100 - k}] a^{100 - k} x^{k}

1 + {(- 1)}^{100 - k} = 2 f o r k = 0, 2, \dots, 100 \Rightarrow 51 t e r m s

1 + {(- 1)}^{100 - k} = 0 f o r k = 1, 3, \dots, 99 \Rightarrow 50 t e r m s

\Rightarrow {(x + a)}^{100} + {(x - a)}^{100} h a s 51 t e r m s .

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

T h a n k s f o r c o r r e c t i o n \dots .

Commented by Tawa1 last updated on 05/May/19

God bless you sir

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