Question

Find the sum of 1+4+13+40+120+....... n terms​

Answer 1

Answer:

one hundred and seventy eight

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

ꜰɪɴᴅ ᴛʜᴇ ꜱᴜᴍ ᴏꜰ 1+4+13+40+120+....... ɴ ᴛᴇʀᴍꜱ​

♣ ᴀɴꜱᴡᴇʀ :

$\sf{Let \:T_{n}\: be\: the\: nth\: term\: and \:S_{n}\: be\: the\: sum\: to\: n\: terms\: of\: the \:given \:series.}$

Thus, we have:

$\sf{S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n} \quad \ldots .(1)}$

Equation (1) can be rewritten as:

$\sf{S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n} \ldots(2)}$

On subtracting (2) from (1), we get:

$\sf{S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n}}$

$\sf{S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n}}$

The sequence of difference between successive terms is 3,9,27,81, ....

We observe that it is a GP with common ratio 3 and first term 3 .

Thus, we have:

$\sf{1+\left[\dfrac{3\left(3^{n-1}-1\right)}{3-1}\right]-T_{n}=0}$

$\sf{\Rightarrow 1+\left[\dfrac{\left(3^{n}-3\right)}{2}\right]-T_{n}=0}$

$\sf{\Rightarrow\left(\dfrac{3^{n}}{2}-\dfrac{1}{2}\right)-T_{n}=0}$

$\sf{\Rightarrow\left(\dfrac{3^{n}}{2}-\dfrac{1}{2}\right)=T_{n}}$

$\sf{\because S_{n}=\sum_{k=1}^{n} T_{k}}$

$\bf{\therefore S_{n}=\sum_{l=1}^{n}\left(\dfrac{3^{k}}{2}-\dfrac{1}{2}\right)}$

$\sf{\Rightarrow S_{n}=\dfrac{1}{2} \sum_{k=1}^{n} 3^{k}-\dfrac{1}{2} \sum_{k=1}^{n} 1}$ \\

$\sf{\Rightarrow S_{n}=\dfrac{1}{2}\left(3+3^{2}+3^{3}+3^{4}+3^{5}+\ldots+3^{n}\right)-\dfrac{n}{2}}$

$\sf{\Rightarrow S_{n}=\dfrac{1}{2}\left[\dfrac{3\left(3^{n}-1\right)}{2}\right]-\dfrac{n}{2}}$

$\sf{\Rightarrow S_{n}=\left(\dfrac{3^{n+1}-3}{4}\right)-\dfrac{n}{2}}$

$\bf{\Rightarrow S_{n}=\left(\dfrac{3^{n+1}-3-2 n}{4}\right)}$