Question

Find the sum to n terms of the series 7+77+777+................

Answer 1

We can approach this question by expanding each term in terms of multiples of 10 and then try to get a general term so that it forms a sequence.

Complete step by step answer:
Now by expanding the given sequence we get,

& 7+77+777+.....n terms \\\ & \Rightarrow \left( 7\times 1 \right)+\left( \left[ 7\times 10 \right]+\left[ 7\times 1 \right] \right)+\left( \left[ 7\times 100 \right]+\left[ 7\times 10 \right]+\left[ 7\times 1 \right] \right)+...nterms \\\ & \Rightarrow \left[ \left( 7\times 1 \right)+\left( 7\times 1 \right)+\left( 7\times 1 \right)...n terms \right]\\\ & +\left[ \left( 7\times 10 \right)+\left( 7\times 10 \right)+.....\left( n-1 \right)terms \right]+\left[ \left( 7\times 100 \right)+...\left( n-2 \right)terms \right] \\\ & \\\ \end{aligned}$$ On observing the first part where 7 is multiplied with ones we can find that 7 is being added n times and in the second part 7 multiplied with tens is being added n-1 times then in the third part where 7 multiplied with hundred being added n-2 times. Similarly this continues till the n terms where 7 multiplied by $${{10}^{n}}$$ will be added one time. So, we can find a generalisation and can expand accordingly. $$\Rightarrow \left( 7+7+7.....ntimes \right)+\left( 70+70........\left( n-1 \right)times \right)+\left( 700+.....\left( n-2 \right)times \right)$$ Now this becomes as follows: $$\Rightarrow \left( 7\times n \right)+\left( 7\times 10\times \left( n-1 \right) \right)+\left( 7\times 100\times \left( n-2 \right) \right)+...........+upto\text{ }n\text{ }terms$$ Here by writing all the n terms together and other constant terms together we get, $$\Rightarrow \left[ \left( 7\times n \right)+\left( 7\times 10\times n \right)+\left( 7\times 100\times n \right)+..... \right]+\left[ \left( 10\times -1 \right)+\left( 100\times -2 \right)+.... \right]$$ Now we can take 7 and n common in the first part and -1 in the second part and then rewrite it as: $$\Rightarrow 7n\left[ 1+10+100+......+{{10}^{n-1}} \right]-\left[ 10+\left( 2\times {{10}^{2}} \right)+.....+\left( n\times {{10}^{n}} \right) \right]$$ As we already know that, A sequence in which the ratio of two consecutive terms is constant is called GP. The constant ratio is called common ratio(r). i. e., $$\dfrac{{{a}_{n+1}}}{{{a}_{n}}}=r..............\left( 1 \right)$$ where n is greater than or equal to one Where terms are in the form of $a, ar, ar^2.....ar^{n-1}$ Sum of n terms of a GP is given by: $${{s}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},if\text{ }r>1................\left( 2 \right)$$ Here r is the common ratio and a is the first term of the sequence. A sequence in which every term is a product of a term of AP and GP is known as AGP series called arithmetic-geometric progression. The series maybe in the form of a, (a+d) r, ......,[a+(n-1)d ]rn-1 Where a is the first term, d is the common difference and r is the common ratio Sum of n terms of AGP series is given by $${{s}_{n}}=\dfrac{a}{1-r}+\dfrac{dr\left( 1-{{r}^{n-1}} \right)}{{{\left( 1-r \right)}^{2}}}-\dfrac{\left\\{ a+\left( n-1 \right)d \right\\}{{r}^{n}}}{1-r},if\text{ }r\ne 1................\left( 3 \right)$$ Here by observing the first part and on comparing the terms we can say that it forms a GP series whereas similarly the second part forms an AGP series. On solving the first part and on comparing it with the equation (1) we get, r=10 and a=1 now by applying these values in the equation (2) we get, $$7n\left[ 1+10+100+......+{{10}^{n-1}} \right]=7n\times {{s}_{n}}..............\left( 4 \right)$$ $$\begin{aligned} & \Rightarrow {{s}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} \\\ & \Rightarrow {{s}_{n}}=\dfrac{1\left( {{10}^{n}}-1 \right)}{10-1} \\\ & \Rightarrow {{s}_{n}}=\dfrac{1}{9}\left( {{10}^{n}}-1 \right) \\\ \end{aligned}$$ Now by substituting this in equation (4) we get first part as: $$\therefore \dfrac{7n}{9}\left( {{10}^{n}}-1 \right)$$ Now by considering the second part and on comparing it with the AGP series we get, a=1, d=1 , r=10 by substituting the respective values on equation (3) we get in the second part as: $$-10\left[ 1+\left( 2\times 10 \right)+.....+\left( n\times {{10}^{n-1}} \right) \right]=-10\times {{s}_{n}}...........(5)$$ $$\begin{aligned} & \Rightarrow {{s}_{n}}=\dfrac{a}{1-r}+\dfrac{dr\left( 1-{{r}^{n-1}} \right)}{{{\left( 1-r \right)}^{2}}}-\dfrac{\left\\{ a+\left( n-1 \right)d \right\\}{{r}^{n}}}{1-r} \\\ & \Rightarrow {{s}_{n}}=\dfrac{1}{1-10}+\dfrac{1\times 10\left( 1-{{10}^{n-1}} \right)}{{{\left( 1-10 \right)}^{2}}}-\dfrac{\left\\{ 1+\left( n-1 \right)1 \right\\}{{10}^{n}}}{1-10} \\\ & \Rightarrow {{s}_{n}}=\dfrac{1}{9}+\dfrac{10\left( 1-{{10}^{n-1}} \right)}{81}+\dfrac{n{{10}^{n}}}{9} \\\ \end{aligned}$$ Now, by substituting this in equation (5) we get, $$\Rightarrow -10\left( \dfrac{1}{9}+\dfrac{10\left( 1-{{10}^{n-1}} \right)}{81}+\dfrac{n{{10}^{n}}}{9} \right)$$ Here, by adding both the parts we can write it as: $$\Rightarrow \dfrac{7n}{9}\left( {{10}^{n}}-1 \right)+-10\left( \dfrac{1}{9}+\dfrac{10\left( 1-{{10}^{n-1}} \right)}{81}+\dfrac{n{{10}^{n}}}{9} \right)$$ Now by writing all the $${{10}^{n}}$$ terms together and constant together we get, $$\Rightarrow \dfrac{7n}{9}\times {{10}^{n}}-\dfrac{10}{81}\times \left( -10\times {{10}^{n-1}} \right)-\dfrac{10n}{9}\times {{10}^{n}}-\dfrac{7n}{9}-\dfrac{10}{9}-\dfrac{100}{81}$$ $$\Rightarrow {{10}^{n}}\left( \dfrac{7n}{9}+\dfrac{10}{81}-\dfrac{10n}{9} \right)-\dfrac{7n}{9}-\dfrac{190}{81}$$ $$\Rightarrow {{10}^{n}}\left( \dfrac{10}{81}-\dfrac{3n}{9} \right)-\dfrac{7n}{9}-\dfrac{190}{81}$$ $$\Rightarrow {{10}^{n}}\left( \dfrac{10-27n}{81} \right)-\left( \dfrac{63n+190}{81} \right)$$ **$$\therefore \text{Hence, 7+77+777+}.......\text{=}{{10}^{n}}\left( \dfrac{10-27n}{81} \right)-\left( \dfrac{63n+190}{81} \right)$$.** **Note:** It is important to write 77 and 777 and so on in the multiples of 10 which helps in getting the GP and AGP series easily. Also it is crucial to find that the sequence is in GP and AGP if not this method it cannot be solved easily manually. While writing the sum of n terms of the series we need to be careful in substituting the values in the respected formula. Identifying that the sequence in the second part is in AGP is one of the crucial step.