If 1 - x2 1 - x2 - x4 1 - x2 - x4 - x6-1 - x2 - x4 - - x2n - -k-0n n-1 akxk- then a0 - a2 - a4 - - - ann-1 equals to

(1 + x^2) (1 + x^2 + x^4)........(1 + x^2 +.........+x^2n) = ∑_k = 0^n(n+1)a_k.x^k 1.2.3.4.............(n + 1) = a_0 + a_1 + a_2 + ........+ a_n(n + 1) nbsp; ...

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Byju's Answer

Standard IX

Mathematics

Inverse of a Matrix

If 1 + x2 1 +...

Question

If $(1 + x^{2}) (1 + x^{2} + x^{4}) (1 + x^{2} + x^{4} + x^{6}) . . . . (1 + x^{2} + x^{4} + . . . . x^{2 n}) = n (n + 1) \sum k = 0 a_{k} x^{k},$ then $a_{0} + a_{2} + a_{4} + . . . . + a_{n (n + 1)}$ equals to

\frac{1}{2} (n + 1)!

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\frac{1}{2} (2 n)!

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(n + 1)!

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{\frac{(n!)}{2}}^{2}

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Solution

The correct option is C $(n + 1)!$
$(1 + x^{2}) (1 + x^{2} + x^{4}) . . . . . . . . (1 + x^{2} + . . . . . . . . . + x^{2 n}) = n (n + 1) \sum k = 0 a_{k} . x^{k}$
$1.2.3.4............. (n + 1) = a_{0} + a_{1} + a_{2} + . . . . . . . . + a_{n (n + 1)}$ ...(i)
$1.2.3.4............. (n + 1) = a_{0} - a_{1} + a_{2} + . . . . . . . . + a_{n (n + 1)}$ ...(ii)
Add (i) and (ii)
$2 {a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{n (n + 1)}} = 2 (n + 1)!$
$a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{n (n + 1)} = (n + 1)!$

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Similar questions

Q. If

(1 + x + x^{2}) (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + . . .

then,

\int \frac{x^{4} - 1}{x^{2} \sqrt{x^{4} + x^{2} + 1}} d x

is equal to

\int \frac{d x}{x^{4} \sqrt{a^{2} + x^{2}}} = \frac{1}{a^{4}} [\frac{1}{x} \sqrt{a^{2} + x^{2}} - \frac{1}{3 x^{3}} {(a^{2} + x^{2})}^{3 / 2}]

. If this is true enter 1, else enter 0.

Q. Let

(1 + x) (1 + x + x^{2}) (1 + x + x^{2} + x^{3}) . . . . . . (1 + x + x^{2} + . . . . + x^{30}) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{465} x^{465}

then sum of

a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{464}

Q. If

{(1 + x + x^{2} + x^{3})}^{5} = 15 \sum k = 0 a_{k} x^{k}

then

7 \sum k = 0 a_{2 k}

is equal to

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If (1+x2)(1+x2+x4)(1+x2+x4+x6)....(1+x2+x4+....x2n)=n(n+1)∑k=0akxk, then a0+a2+a4+....+an(n+1) equals to

The correct option is C (n+1)!(1+x2)(1+x2+x4)........(1+x2+.........+x2n)=n(n+1)∑k=0ak.xk 1.2.3.4.............(n+1)=a0+a1+a2+........+an(n+1) ...(i) 1.2.3.4.............(n+1)=a0–a1+a2+........+an(n+1) ...(ii) Add (i) and (ii) 2{a0+a2+a4+........+an(n+1)}=2(n+1)! a0+a2+a4+........+an(n+1)=(n+1)!

If $(1 + x^{2}) (1 + x^{2} + x^{4}) (1 + x^{2} + x^{4} + x^{6}) . . . . (1 + x^{2} + x^{4} + . . . . x^{2 n}) = n (n + 1) \sum k = 0 a_{k} x^{k},$ then $a_{0} + a_{2} + a_{4} + . . . . + a_{n (n + 1)}$ equals to