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7.Binomial Theorem
easy
${(1 + x + {x^2} + {x^3})^5}$ ના વિસ્તરણમાં $x$ ની યુગ્મ ઘાતકના સહગુણકનો સરવાળો મેળવો.
A
$256$
B
$128$
C
$512$
D
$64$
Solution
(c) ${(1 + x + {x^2} + {x^3})^5} = {(1 + x)^5}{(1 + {x^2})^5}$
$ = (1 + 5x + 10{x^2} + 10{x^3} + 5{x^4} + {x^5})$$ \times (1 + 5{x^2} + 10{x^4} + 10{x^6} + 5{x^8} + {x^{10}})$
Therefore the required sum of coefficients
$ = (1 + 10 + 5){.2^5} = 16 \times 32 = 512$
Note : ${2^n} = {2^5}$= Sum of all the binomial coefficients in the $2^{nd}$ bracket in which all the powers of $x$ are even.
Standard 11
Mathematics
