If 1 + {x^4} + {x^5} = sumlimits_{i = 0} | vedclass

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Question

$1 + {x^4} + {x^5} = \sum\limits_{i = 0}^5 {{a_i}} {(1 + x)^i}$

$ = {a_0} + {a_1}{(1 + x)^1} + {a_2}{(1 + x)^2} + {a_3}{(1 + x)^\beta }$

$ + {a_

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

$1 + {x^4} + {x^5} = \sum\limits_{i = 0}^5 {{a_i}} {(1 + x)^i}$

$ = {a_0} + {a_1}{(1 + x)^1} + {a_2}{(1 + x)^2} + {a_3}{(1 + x)^\beta }$

$ + {a_4}{(1 + x)^4} + {a_5}{(1 + x)^5}$

$ \Rightarrow 1 + {x^4} + {x^5}$

$ = {a_0} + {a_1}(1 + x) + {a_2}\left( {1 + 2x + {x^2}} \right)$

$ + {a_3}\left( {1 + 3x + 3{x^2} + {x^3}} \right)$

$ + {a_4}\left( {1 + 4x + 6{x^2} + 4{x^3} + {x^4}} \right)$

$ + {a_5}(1 + 5x + 10{x^2} + 10{x^3} + 5{x^4} + {x^5})$

$ \Rightarrow 1 + {x^4} + {x^5}$

$ = {a_0} + {a_1} + {a_1}x + {a_2} + 2{a_2}x + {a_2}{x^2}$

$ + {a_3} + 3{a_3}x + 3{a_3}{x^2} + {a_3}{x^3}$

$ + {a_4} + 4{a_4}x + 6{a_4}{x^2} + 4{a_4}{x^3} + {a_4}{x^4}$

$ + {a_5} + 5{a_5}x + 10{a_5}{x^2} + 10{a_5}{x^3} + 5{a_5}{x^4} + {a_5}{x^5}$

$\Rightarrow 1+x^{4}+x^{5}$

$\left(a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)$

$=+x\left(a_{1}+2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}\right)$

$+x^{2}\left(a_{2}+3 a_{3}+6 a_{4}+10 a_{5}\right)+x^{3}\left(a_{3}+4 a_{4}+10 a_{5}\right)$

$+x^{4}\left(a_{4}+5 a_{5}\right)+x^{5}\left(a_{5}\right)$

On comparing the like coefficients, we get

$\boxed{{a_5} = 1}.........(1)$

$\boxed{{a_4} + 5{a_5} = 1}.........(2)$

$\boxed{{a_3} + 4{a_4} + 10{a_5} = 0}.........(3)$

and $\boxed{{a_2} + 3{a_3} + 6{a_4} + 10{a_5} = 0}.........(4)$

from$(1)$ and $(2)$, we get

$\boxed{{a_4} =  - 4}.........(5)$

from $(1),(3)$  and  $(5),$ we get

$\boxed{{a_3} =  + 6}.........(6)$

Now, from $(1)$, $(5)$ and $(6)$, we get 

$a_{2}=-4$

If $1 + {x^4} + {x^5} = \sum\limits_{i = 0}^5 {{a_i}\,(1 + {x})^i,} $ for all $x$ in $R,$ then $a_2$ is

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