Question

find the sum to n terms of 1+5+12+22+35+...

Answer 1

The given series is 1+5+12+22+35+.....
difference are 4, 7, 10, 13,........
Therefore, it is in AP.
Let, Tn=an^2+bn+c
T1=a+b+c=1..........(1)
T2=4a+2b+c=5.........(2)
T3=9a+3b+c=12..........(3)
subtracting eq.(1) from (2) we get,
4a+2b+c=5
a + b +c=1
_____________
3a+b =4..............(4)
Now, subtracting eq.(2) from (3) we get,
9a+3b+c=12
4a+2b+c= 5
______________
5a+b =7..............(5)
Now, subtracting eq.(4) from (5) we get ,
5a+b=7
3a+b=4
_________
2a =3
a=3/2
putting the value of 'a' in eq.(4) we get,
3×3/2+b =4
9/2+b =4
b=4-9/2
b=-1/2
Now putting the value of 'a' and 'b' in eq. (1) we get,
3/2-1/2+c=1
2/2+c=1
1+c=1
c=1-1
c=0
substituting the value of 'a', 'b' and 'c' in Tn we get,
Tn= 3n^2- n/2
Sn=  ΣTn
=  Σ(3n^2-n/2)
= 3/2 n(n+1)(2n+1)/6 - 1/2 n(n+1)/2
= 1/4 [n(n+1)] [2n+1-1]
= n(n+1)n/2
= n^2 (n+1)/2

Answer 2

Step-by-step explanation:

let Tn be the nth term

Tn = an²+bn+c

T1 = a+b+c = 1

T2 = 4a+2b+c = 5

T3 = 9a+3b+c = 12

T2-T1 => 3a+b = 4

T3-T2 => 5a+b = 7

(5a+b)-(3a+b)=7-4

2a=3

a = 3/2

b = -1/2

c = 0

put the values of a, b and c in the equation of Tn

Tn = 3/2n²-1/2n

Sn = £(3/2n²-1/2n)

(after putting the values of sigma)

n²(n+1)

thanks...