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Ncert exemplar solutions

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n^2-1 is divisible by 8, if n is

Let p=n^2-1

Case I: Let n is even, then n = 2k.

\therefore p=(2k)^2-1 \\ p=4k^2-1

Let k = 0, then p = 4(0.0) – 1 = -1, which is not divisible by 8

k = 2, then p = 4(2.2) – 1 = 15, which is not divisible by 8

k = 4, then p = 4(4.4) = 64 – 1 = 6, which is not divisible by 8

So, n cannot be even integer.

Case II: Let n is odd then n = 2k + 1

\begin{array}{l} p=(2 k+1)^{2}-1 \\ =4 k^{2}+1+4 k-1 \\ p=4 k(k+1) \end{array}

Let k = 1, p = 4(1) [1+1] = 8 is divisible by 8

k = 3 p = 4×3 (3+1) = 48 = 8 × 6, is divisible by 8

k = 5 p = 4(5) (5 + 1) = 120, = 8 × 15 is divisible by 8

So, n² – 1 is divisible by 8 if n is odd number.

n^2-1 is divisible by 8, if n is

Let p=n^2-1

Case I: Let n is even, then n = 2k.

\therefore p=(2k)^2-1 \\ p=4k^2-1

Let k = 0, then p = 4(0.0) – 1 = -1, which is not divisible by 8

k = 2, then p = 4(2.2) – 1 = 15, which is not divisible by 8

k = 4, then p = 4(4.4) = 64 – 1 = 6, which is not divisible by 8

So, n cannot be even integer.

Case II: Let n is odd then n = 2k + 1

\begin{array}{l} p=(2 k+1)^{2}-1 \\ =4 k^{2}+1+4 k-1 \\ p=4 k(k+1) \end{array}

Let k = 1, p = 4(1) [1+1] = 8 is divisible by 8

k = 3 p = 4×3 (3+1) = 48 = 8 × 6, is divisible by 8

k = 5 p = 4(5) (5 + 1) = 120, = 8 × 15 is divisible by 8

So, n² – 1 is divisible by 8 if n is odd number.

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