10. `n^3-n^2` type series
1. Find next 3 numbers in the sequence `6,21,91,441,2275,12201,67171`
Each number in the series is of the form `1^n+2^n+3^n+4^n+5^n+6^n`
`1^0+2^0+3^0+4^0+5^0+6^0=1+1+1+1+1+1=6`
`1^1+2^1+3^1+4^1+5^1+6^1=1+2+3+4+5+6=21`
`1^2+2^2+3^2+4^2+5^2+6^2=1+4+9+16+25+36=91`
`1^3+2^3+3^3+4^3+5^3+6^3=1+8+27+64+125+216=441`
`1^4+2^4+3^4+4^4+5^4+6^4=1+16+81+256+625+1296=2275`
`1^5+2^5+3^5+4^5+5^5+6^5=1+32+243+1024+3125+7776=12201`
`1^6+2^6+3^6+4^6+5^6+6^6=1+64+729+4096+15625+46656=67171`
So required number are
`1^7+2^7+3^7+4^7+5^7+6^7=1+128+2187+16384+78125+279936=376761`
`1^8+2^8+3^8+4^8+5^8+6^8=1+256+6561+65536+390625+1679616=2142595`
`1^9+2^9+3^9+4^9+5^9+6^9=1+512+19683+262144+1953125+10077696=12313161`
Answer : `6,21,91,441,2275,12201,67171,376761,2142595,12313161`
`:.` The next 3 number for given series `6,21,91,441,2275,12201,67171` are `376761,2142595,12313161`
Solution-1
2. Find next 3 numbers in the sequence `6,14,36,98`
Each number in the series is of the form `1^n+2^n+3^n`
`1^1+2^1+3^1=1+2+3=6`
`1^2+2^2+3^2=1+4+9=14`
`1^3+2^3+3^3=1+8+27=36`
`1^4+2^4+3^4=1+16+81=98`
So required number are
`1^5+2^5+3^5=1+32+243=276`
`1^6+2^6+3^6=1+64+729=794`
`1^7+2^7+3^7=1+128+2187=2316`
Answer : `6,14,36,98,276,794,2316`
`:.` The next 3 number for given series `6,14,36,98` are `276,794,2316`
Solution-1
3. Find next 3 numbers in the sequence `4,18,48,100,180,294,448`
Series is of the form `n^3-n^2` or `(n-1)*n^2`
`2^3-2^2=8-4=4` or `(2-1)*2^2=1*4=4`
`3^3-3^2=27-9=18` or `(3-1)*3^2=2*9=18`
`4^3-4^2=64-16=48` or `(4-1)*4^2=3*16=48`
`5^3-5^2=125-25=100` or `(5-1)*5^2=4*25=100`
`6^3-6^2=216-36=180` or `(6-1)*6^2=5*36=180`
`7^3-7^2=343-49=294` or `(7-1)*7^2=6*49=294`
`8^3-8^2=512-64=448` or `(8-1)*8^2=7*64=448`
So required number are
`9^3-9^2=729-81=648` or `(9-1)*9^2=8*81=648`
`10^3-10^2=1000-100=900` or `(10-1)*10^2=9*100=900`
`11^3-11^2=1331-121=1210` or `(11-1)*11^2=10*121=1210`
Answer : `648,900,1210`
`:.` The next 3 number for given series `4,18,48,100,180,294,448` are `648,900,1210`
Solution-1
4. Find next 3 numbers in the sequence `1,5,19,65`
Series is of the form `3^n-2^n`
`3^1-2^1=3-2=1`
`3^2-2^2=9-4=5`
`3^3-2^3=27-8=19`
`3^4-2^4=81-16=65`
So required number are
`3^5-2^5=243-32=211`
`3^6-2^6=729-64=665`
`3^7-2^7=2187-128=2059`
Answer : `211,665,2059`
`:.` The next 3 number for given series `1,5,19,65` are `211,665,2059`
Solution-1
This material is intended as a summary. Use your textbook for detail explanation.
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