Question

If $n^{th}$ term of the series 1 + 5 + 12 + 22 + 35 .........can be written as $T_n$ = a$n^2$ + bn + c Find the sum of the 16 terms of the series.

• A. 2176
• B. 7076
• C. 3176
• D. 3000

Answer 1

The correct option is A

2176

Given

$T_n$ = a$n^2$ + bn + c-----------------(1)

putting n = 1,2,3,.......so on

We get,

$T_1$ = a + b + c = 1----------------------(2)

$T_2$ = 4a + 2b + c = 5------------------(3)

$T_3$ = 9a + 3b + c = 12------------------(4)

Multiply 4 in equation 2 and subtract equation 3 from it.

2b + 3c = -1------------(5)

Multiply 9 in equation 2 and subtract equation 4 from it.

6b + 8c = -3------------(6)

Multiply 3 in equation 5 and subtract equation 6 from it.

c = 0

Substitute c in equation 5

We get, b = -$\frac{1}{2}$

Substitute b and c in equation 2

We get a + b + c = 1, a -$\frac{1}{2}$ + 0 = 1

a = $\frac{3}{2}$

So,

a = $\frac{3}{2}$, b = -$\frac{1}{2}$ and c = 0

Substituting these values in equation 1,we get,

$T_n$ = $\frac{3}{2}$ $n^2$ - $\frac{1}{2}$ n = $\frac{1}{2}$ $(3n^2-n)$

sum of the given series.

$S_n={\sum }_{r=1}^n{T}_n$ = $\frac{1}{2}$ ${\sum }_{r=1}^n(3n^2-n)$

= $\frac{1}{7}$ $\left[3.\frac{n}{6}\right(n+1\left)\right(2n+1)-\frac{n(n+1)}{2}]$

= $\frac{1}{2}$. $\frac{1}{2}$ n(n + 1)[2n + 1.1]

= $\frac{1}{4}$ n(n + 1) $\times$ 2n

= $\frac{n^2(n+1)}{2}$

= Substitute n = 16

Sum of the 16 terms $S_{16}$ = $\frac{16\times 16\times 17}{2}$ = 2176