Series-3 (NCERT)

Find sum to n terms of the following series:

1² + (1²+2²) + (1²+2²+3²) +……….

(1) n(n+1)(n+2)²/12 

(2) n²(n+1)(n+2)/12  

(3) n(n+1)²(n+2)/12  

(4) n(n+1)(2n+1)/6  

(5) n²(n+1)²/4

Answer follows here:

Solution:

The formulae to be used here are:

Sum of ‘n’ consecutive integers starting from 1 = 1+2+3+…..+n =  ∑n = n(n+1)/2

Sum of squares of ‘n’ consecutive integers starting from 1

= 1²+2²+3²+…..+n² = ∑n² = n(n+1)(2n+1)/6

Sum of cubes of ‘n’ consecutive integers starting from 1

= 1³+2³+3³+…..+n³ = ∑n³ = n²(n+1)²/4

Given series is:

1² + (1²+2²) + (1²+2²+3²) +……….

n^th term of the series = t_n = 1²+2²+3²+…..+n²  = ∑n² = n(n+1)(2n+1)/6

Sum to n terms = ∑t_n = ∑n(n+1)(2n+1)/6

= 1/6 ∑(n²+n)(2n+1)

= 1/6 ∑(2n³+ n²+2n²+n)

= 1/6 ∑(2n³+3n²+n)

= 1/3 ∑n³ +  1/2 ∑n² +  1/6 ∑n

= 1/3 (n²(n+1)²/4) + ½ (n(n+1)(2n+1)/6) + 1/6 (n(n+1)/2)

= n(n+1)/12 {n(n+1)+ (2n+1) + 1}

= n(n+1)/12 {n²+n+2n+2}

= n(n+1)/12 {n(n+1)+2(n+1)}

= n(n+1)/12 {(n+1)(n+2)}

= n(n+1)²(n+2)/12

Answer (3)


Short cut method:

As “None of these” option is not given, we can safely check the answer options by substituting value of ‘n’:

For n =1, LHS = 1² = 1

For RHS answer options can be checked one by one:

Option(1) ----   1(1+1)(1+2)²/12 = 1*2*9/12 ≠ 1 => this option is wrong

Option(2) ----   1²(1+1)(1+2)/12 = 1*2*3/12 ≠ 1 => this option is wrong

Option(3) ----   1(1+1)²(1+2)/12 = 1*4*3/12 = 1 => this option is correct

Option(4) ----   1(1+1)(2(1)+1)/6 = 1*2*3/6 = 1 => this option is correct

Option(5) ----   1²(1+1)²/4 = 1*4/4 = 1 => this option is correct

Left out options are (3),(4) and (5).

Now we check for n=2:

LHS = 1² + (1²+2²)= 1+5 = 6

Option(3) ----   2(2+1)²(2+2)/12 = 2*9*4/12 = 6 => this option is correct

Option(4) ----   2(2+1)(2(2)+1)/6 = 2*3*5/6 ≠ 6 => this option is wrong

Option(5) ----   2²(2+1)²/4 = 4*9/4 ≠ 6 => this option is wrong

The only left out option is (3)