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`        please help me to solve this ∑ (k^2+k+1)k! where k=1 to 100 `
7 years ago

```							Dear Shashank,

n kk=1
= 1 + 2 + 3 + 4 + .. + n
= (n2 + n) / 2= (1/2)n2 + (1/2)n
sum of 1stn integers

n k 2k=1
= 1 + 4 + 9 + 16 + .. + n2
= (1/6)n(n+1)(2n+1)= (1/3)n3 + (1/2)n2 + (1/6)n
sum of 1stn squares

n k 3k=1
= 1 + 8 + 27 + 64 + .. + n3
= (1/4)n4 + (1/2)n3 + (1/4)n2
sum of 1stn cubes

n k 4k=1
= 1 + 16 + 81 + 256 + .. + n4
= (1/5)n5 + (1/2)n4 + (1/3)n3 - (1/30)n

n k 5k=1
= 1 + 32 + 243 + 1024 + .. + n5
= (1/6)n6 + (1/2)n5 + (5/12)n4 - (1/12)n2

n k 6k=1
= 1 + 64 + 729 + 4096 + .. + n6
= (1/7)n7 + (1/2)n6 + (1/2)n5 - (1/6)n3 + (1/42)n

n k 7k=1
= 1 + 128 + 2187 + 16384 + .. + n7
= (1/8)n8 + (1/2)n7 + (7/12)n6 - (7/24)n4 + (1/12)n2

n k 8k=1
= 1 + 256 + 6561 + 65536 + .. + n8
= (1/9)n9 + (1/2)n8 + (2/3)n7 - (7/15)n5 + (2/9)n3 - (1/30)n

n k 9k=1
= 1 + 512 + 19683 + 262144 + .. + n9
= (1/10)n10 + (1/2)n9 + (3/4)n8 - (7/10)n6 + (1/2)n4 - (3/20)n2

n k 10k=1
= 1 + 1024 + 59049 + 1048576 + .. + n10
= (1/11)n11 + (1/2)n10 + (5/6)n9 - n7 + n5 - (1/2)n3 + (5/66)n Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….
So start the brain storming…. become a leader with Elite Expert League ASKIITIANS
Thanks
Aman Bansal

```
7 years ago
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