7.Binomial Theorem
hard

बहुपद $(x - 1)(x - 2)(x - 3).............(x - 100),$ में ${x^{99}}$ का गुणांक होगा  

A

$5050$

B

$-5050$

C

$100$

D

$99$

Solution

$(x – 1)(x – 2)(x – 3)….(x – 100)$

पदों की संख्या $= 100$; 

 $(x – 1)(x – 2)(x – 3)…(x – 100)$ में  $x^{99}$ का गुणांक  =

= $( – 1 – 2 – 3 – …… – 100)$ = $ – (1 + 2 + ….. + 100)$     

= $ – \frac{{100 \times 101}}{2}$ $= -5050$

Standard 11
Mathematics

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