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Dev Kumar Aidaswani 6 years, 8 months ago

We first take P (n)=1 Then, P (n)=1 (1+1)(1+2) =1 (2)(3) P (n)=6 Therefore 6is divisible by3 So,P (n)is true for 1 Now let us assume k (k+1)(k+2)=3x Now let p (n)=k+1 So, K+1 (K+2)(K+3) SO IF K(K+1)(K+2) IS DIVISIBLE BY 3 SO EITHER ONE OF THEM IS DIVISIBLE BY 3 IF K IS MULTIPLE OF 3 SO IN K+1 (K+2)(K+3) K+3 IS MULTIPLE OF 3 SO THE ABOVE NO. IS DIVISIBLE BY 3 SIMILARILY THE CASE APPLIED WITH K+1 AND K+2 WE GET IN ALL CASES 3 IS DIVISIBLE THEREFORE K (K+1)(K+2) IS DIVISIBLE BY 3 HENCE PROVED
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Yogita Ingle 6 years, 8 months ago

(a - b)2 = a2 - 2ab + b2

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Ravishankar Sahu 6 years, 8 months ago

{1},{2},{3 },{4},{5},{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5},{1,2,3},{2,3,4},{3,4,5},{1,3,4},{1,4,5},{2,4,5},{1,3,5},{1,2,4},{2,3,5},{1,2,5},{1,2,3,4},{2,3,4,5},{1,3,4,5}{1,2,4,5},{1,2,3,5},{1,2,3,4,5},{ø}.
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Mohd Zubair 6 years, 8 months ago

this is example
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Sarthak Agarwal 6 years, 8 months ago

4- 6 marker questions

Sarthak Agarwal 6 years, 8 months ago

Yes now there are 20-1 marker questions ,6 2marker, 6 4markers and

Prabh Singh 6 years, 8 months ago

No pattern has been changed till today.
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Ujjwal Mishra 6 years, 8 months ago

(k-1)x^2=kx-1 This can be written as; (k-1)x^2-(kx-1)=0 (k-1)x^2-kx+1=0 Put x= -3 as this is a zero of polynomial: (k-1)(-3)^2-k(-3)+1=0 (k-1)(9)+3k+1=0 9k-9+3k+1=0 9k+3k-9+1=0 12k-8=0 12k=8 k=8/12 k=2/3 So, the value of k is 2/3 so that one of the zero of this polynomial is -3. #Hope this may be helpful for you...
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Ujjwal Mishra 6 years, 8 months ago

2n^2+1=5 We need to find the value of n here; So, 2n^2+1=5 2n^2=5-1=4 2n^2=4 n^2=4/2=2 n=+-√2 n=+√2 or -√2
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Manoj Kunwar 6 years, 8 months ago

Addition is the sum of two or more than two numbers. Example - 2+2=4 , 15+7= 22 , 0+0=0
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