Question

for any positive integer n, prove that n^3-n is divisable by 6.

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Answer 1

Answer:

For any Positive integer n, prove that n3−n divisible by 6. The condition for any number to be divisible by 6 is that the number must be individually divisible by 3 and 2. ... If n=3p+2, then n+1=3p+2+1=3(p+1). The number is divisible by 3.

Step-by-step explanation:

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Answer 2

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ᴛʜᴇʀᴇ ᴀʀᴇ ᴛᴡᴏ ᴍᴇᴛʜᴏᴅꜱ ᴛᴏ ꜱᴏʟᴠᴇ ᴛʜᴇ ᴘʀᴏʙʟᴇᴍ ᴡʜɪᴄʜ ᴀʀᴇ ᴅɪꜱᴄᴜꜱꜱᴇᴅ ʙᴇʟᴏᴡ.

ᴍᴇᴛʜᴏᴅ 1:

ʟᴇᴛ ᴜꜱ ᴄᴏɴꜱɪᴅᴇʀ

ᴀ = ɴ3 – ɴ

ᴀ = ɴ (ɴ2 – 1)

ᴀ = ɴ (ɴ + 1)(ɴ – 1)

ᴀꜱꜱᴜᴍᴛɪᴏɴꜱ:

1. ᴏᴜᴛ ᴏꜰ ᴛʜʀᴇᴇ (ɴ – 1), ɴ, (ɴ + 1) ᴏɴᴇ ᴍᴜꜱᴛ ʙᴇ ᴇᴠᴇɴ, ꜱᴏ ᴀ ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2.

2. (ɴ – 1) , ɴ, (ɴ + 1) ᴀʀᴇ ᴄᴏɴꜱᴇᴄᴜᴛɪᴠᴇ ɪɴᴛᴇɢᴇʀꜱ ᴛʜᴜꜱ ᴀꜱ ᴘʀᴏᴠᴇᴅ ᴀ ᴍᴜꜱᴛ ʙᴇ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 3.

ꜰʀᴏᴍ (1) ᴀɴᴅ (2) ᴀ ᴍᴜꜱᴛ ʙᴇ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2 × 3 = 6

ᴛʜᴜꜱ, ɴ³ – ɴ ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 6 ꜰᴏʀ ᴀɴʏ ᴘᴏꜱɪᴛɪᴠᴇ ɪɴᴛᴇɢᴇʀ ɴ.

ᴍᴇᴛʜᴏᴅ 2:

ᴡʜᴇɴ ᴀ ɴᴜᴍʙᴇʀ ɪꜱ ᴅɪᴠɪᴅᴇᴅ ʙʏ 3, ᴛʜᴇ ᴘᴏꜱꜱɪʙʟᴇ ʀᴇᴍᴀɪɴᴅᴇʀꜱ ᴀʀᴇ 0 ᴏʀ 1 ᴏʀ 2.

∴ ɴ = 3ᴘ ᴏʀ 3ᴘ + 1 ᴏʀ 3ᴘ + 2, ᴡʜᴇʀᴇ ʀ ɪꜱ ꜱᴏᴍᴇ ɪɴᴛᴇɢᴇʀ.

ᴄᴀꜱᴇ 1: ᴄᴏɴꜱɪᴅᴇʀ ɴ = 3ᴘ

ᴛʜᴇɴ ɴ ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 3.

ᴄᴀꜱᴇ 2: ᴄᴏɴꜱɪᴅᴇʀ ɴ = 3ᴘ + 1

ᴛʜᴇɴ ɴ – 1 = 3ᴘ + 1 –1

⇒ ɴ -1 = 3ᴘ ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 3.

ᴄᴀꜱᴇ 3: ᴄᴏɴꜱɪᴅᴇʀ ɴ = 3ᴘ + 2

ᴛʜᴇɴ ɴ + 1 = 3ᴘ + 2 + 1

⇒ ɴ+1 = 3ᴘ + 3

⇒ ɴ+1  = 3(ᴘ + 1) ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 3.

ꜱᴏ, ᴡᴇ ᴄᴀɴ ꜱᴀʏ ᴛʜᴀᴛ ᴏɴᴇ ᴏꜰ ᴛʜᴇ ɴᴜᴍʙᴇʀꜱ ᴀᴍᴏɴɢ ɴ, ɴ – 1 ᴀɴᴅ ɴ + 1 ɪꜱ ᴀʟᴡᴀʏꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 3.

⇒ ɴ (ɴ – 1) (ɴ + 1) ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 3.

ꜱɪᴍɪʟᴀʀʟʏ, ᴡʜᴇɴ ᴀ ɴᴜᴍʙᴇʀ ɪꜱ ᴅɪᴠɪᴅᴇᴅ ʙʏ 2, ᴛʜᴇ ᴘᴏꜱꜱɪʙʟᴇ ʀᴇᴍᴀɪɴᴅᴇʀꜱ ᴀʀᴇ 0 ᴏʀ 1.

∴ ɴ = 2Q ᴏʀ 2Q + 1, ᴡʜᴇʀᴇ Q ɪꜱ ꜱᴏᴍᴇ ɪɴᴛᴇɢᴇʀ.

ᴄᴀꜱᴇ 1: ᴄᴏɴꜱɪᴅᴇʀ ɴ = 2Q

ᴛʜᴇɴ ɴ ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2.

ᴄᴀꜱᴇ 2: ᴄᴏɴꜱɪᴅᴇʀ ɴ = 2Q + 1

ᴛʜᴇɴ ɴ–1 = 2Q + 1 – 1

ɴ – 1 = 2Q ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2 ᴀɴᴅ

ɴ + 1 = 2Q + 1 + 1

ɴ +1 = 2Q + 2

ɴ+1= 2 (Q + 1) ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2.

ꜱᴏ, ᴡᴇ ᴄᴀɴ ꜱᴀʏ ᴛʜᴀᴛ ᴏɴᴇ ᴏꜰ ᴛʜᴇ ɴᴜᴍʙᴇʀꜱ ᴀᴍᴏɴɢ ɴ, ɴ – 1 ᴀɴᴅ ɴ + 1 ɪꜱ ᴀʟᴡᴀʏꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2.

∴ ɴ (ɴ – 1) (ɴ + 1) ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2.

ꜱɪɴᴄᴇ, ɴ (ɴ – 1) (ɴ + 1) ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 2 ᴀɴᴅ 3.

ᴛʜᴇʀᴇꜰᴏʀᴇ, ᴀꜱ ᴘᴇʀ ᴛʜᴇ ᴅɪᴠɪꜱɪʙɪʟɪᴛʏ ʀᴜʟᴇ ᴏꜰ 6, ᴛʜᴇ ɢɪᴠᴇɴ ɴᴜᴍʙᴇʀ ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ ꜱɪx.

ɴ3 – ɴ = ɴ (ɴ – 1) (ɴ + 1) ɪꜱ ᴅɪᴠɪꜱɪʙʟᴇ ʙʏ 6.

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