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  • If <img alt="1+x^{4}+x^{5}=\sum_{i=0}^{5}a_{i}\left ( 1+x \right )^{i}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%201&plus;x%5E%7B4%7D&plus;x%5E%7B5%7D%3D%5Csum_%

If 1+x^{4}+x^{5}=\sum_{i=0}^{5}a_{i}\left ( 1+x \right )^{i},for all  x in R, then a2 is :

Option: 1

- 4


Option: 2

6


Option: 3

-8


Option: 4

10


Answers (1)

best_answer

Using

If \sum_{i=0}^{5} a_{i}\left ( 1+ x \right )^{i} = 1 + x^{4} + x^{5}

on expansion, 

1 + a_{1} (1+x)^1+ a_{2}(1+x)^2 + a_{3}(1+x)^3 + a_{4}(1+x)^4+ a_{5}(1+x)^5=1+x^4+x^5

Now differentiate

a_{1}(1+x)^0 +2 a_{2}(1+x)^1 + 3a_{3}(1+x)^2 + 4a_{4}(1+x)^3+ 5a_{5}(1+x)^4=4x^3+5x^4

a_{1} +2 a_{2}(1+x)^1 + 3a_{3}(1+x)^2 + 4a_{4}(1+x)^3+ 5a_{5}(1+x)^4=4x^3+5x^4

Differentiating again

2 a_{2}(1+x)^0 + 6a_{3}(1+x)^1 + 12a_{4}(1+x)^2+ 20a_{5}(1+x)^3=12x^3+20x^4

2 a_{2} + 6a_{3}(1+x)^1 + 12a_{4}(1+x)^2+ 20a_{5}(1+x)^3=12x^2+20x^3

Now put x=-1

2 a_{2} + 6a_{3}(1-1)^1 + 12a_{4}(1-1)^2+ 20a_{5}(1-1)^3=12(-1)^2+20(-1)^3

2 a_{2} + 0=12-20

a_{2}= -4

Posted by

Nehul

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