If P k -1- x k -1-1- x- the number of terms in the product P 1- P 2- P n isA- nn-1-2B- n2-n-2C- n2-n-2-2D- n2-n-2-2

The correct option is D P_k=1 x^k+1/1 x P_1P_2P_3....P_n =(1 x^2)(1 x^3)(1 x^4)....(1 x^n+1)/1 x^n No. of terms = 1 + max. power of x =1+ n(n+1)/2=n^2+n+2/2 ...

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Byju's Answer

Standard XI

Mathematics

Proof by mathematical induction

If P k =1- x ...

Question

If $P_{k} = \frac{1 - x^{k + 1}}{1 - x}$ , the number of terms in the product $P_{1} . P_{2} . . . . P_{n}$ is

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Solution

The correct option is D
$P_{k} = \frac{1 - x^{k + 1}}{1 - x}$
$P_{1} P_{2} P_{3} . . . . P_{n} = \frac{(1 - x^{2}) (1 - x^{3}) (1 - x^{4}) . . . . (1 - x^{n + 1})}{1 - x^{n}}$
No. of terms = 1 + max. power of $x = 1 + \frac{n (n + 1)}{2} = \frac{n^{2} + n + 2}{2}$

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If Pk=1−xk+11−x, the number of terms in the product P1.P2....Pn is

The correct option is D Pk=1−xk+11−x P1P2P3....Pn=(1−x2)(1−x3)(1−x4)....(1−xn+1)1−xn No. of terms = 1 + max. power of x=1+n(n+1)2=n2+n+22

If $P_{k} = \frac{1 - x^{k + 1}}{1 - x}$ , the number of terms in the product $P_{1} . P_{2} . . . . P_{n}$ is

The correct option is D
$P_{k} = \frac{1 - x^{k + 1}}{1 - x}$
$P_{1} P_{2} P_{3} . . . . P_{n} = \frac{(1 - x^{2}) (1 - x^{3}) (1 - x^{4}) . . . . (1 - x^{n + 1})}{1 - x^{n}}$
No. of terms = 1 + max. power of $x = 1 + \frac{n (n + 1)}{2} = \frac{n^{2} + n + 2}{2}$