(1+1*2)+(1+2*5)+(1+3*10)+....+(1+10*101), Find the sum of the following series

Question

show the steps if possible

Answer 1

Question: Find the sum of the following series: (1+1*2)+(1+2*5)+(1+3*10)+....+(1+10*101).

The nth term of the series is given by

tn = 1 + an*bn,

where

{a} = {1,2,3,4,5,...,10}, i.e. an = n

{b} = {2,5,10, ... ,101}

Closer inspection of the elements of {b} show these element as { 1^2 + 1, 2^2 + 1, 3^2 + 1, ... ,10^2 + 1]

i.e. bn = n^2 + 1

Hence tn = 1 + an*bn = 1 + n*(n^2 + 1)

tn = 1 + n + n^3

And sum of these term is is given by

Sn = Sum[k=1..n] (1 + n + n^3) = Sum[k=1..n] (1) + Sum[k=1..n] (n) + Sum[k=1..n] (n^3)

Sn = n + (1/2)n(n+1) + (1/4)n^2(n+1)^2

For n = 10,

S10 = 10 + 5*11 + (5*11)^2 = 10 + 55 + 3025 = 3090

Sum = 3090

Comments

thnx.

Answer 2

We Know,

tn = 1 + an*bn,

where

{a} = {1,2,3,4,5,...,10}, So, an = n                       {b} = {2,5,10, ... ,101}

Nearest elements of {b} show these element as { 1^2 + 1, 2^2 + 1, 3^2 + 1, ... ,10^2 + 1]

Thus, bn = n^2 + 1

Hence tn = 1 + an*bn = 1 + n*(n^2 + 1)

We get that,tn = 1 + n + n^3

And sum of these term is indicate by

Sn = Sum[k=1..n] (1 + n + n^3) = Sum[k=1..n] (1) + Sum[k=1..n] (n) + Sum[k=1..n] (n^3)

Sn = n + (1/2)n(n+1) + (1/4)n^2(n+1)^2

When n = 10,

S10 = 10 + 5*11 + (5*11)^2 = 10 + 55 + 3025 = 3090{ans}                       Safal Das Biswas Class X Savm

Answer 3

Okay so you add 1+1 and that equals 2 multiply that by 2 and that gives you 4..... add 1+2 and that gives you 3 to 4..... multiply 4 by the answer you got when you multiplied 3 and 4...... then add 1+3 and that gives you 4..... multiply that by 10 and that gives you 40..... then add 1+10 and that gives you 11.... then multiply 11 by 101 and that gives you 1,111. So your answer should be 1,111.