Question

There are n A.P’s whose common differences are 1, 2, 3, ……, n respectively, the first term of each being unity. Prove that sum of their ${{n}^{th}}$ term is $\dfrac{1}{2}n\left( {{n}^{2}}+1 \right)$.

Answer 1

Write the terms for each A.P. and find their ${{n}^{th}}$ terms by using the formula: - ${{T}_{n}}=a+\left( n-1 \right)d$, where ${{T}_{n}}$ is the ${{n}^{th}}$ term, ‘a’ is the first term, ‘d’ is the common difference and ‘n’ is the number of terms. Form a general sequence with the ${{n}^{th}}$ term of each A.P. and add them using the formula: - ${{S}_{n}}=\dfrac{n}{2}\times$ (first term + last term). Here, ${{S}_{n}}$ is the sum of ‘n’ terms of an A.P.

Complete step by step answer:
Here, we have been provided with ‘n’ A.P’s whose common differences are 1, 2, 3, ….., n respectively and the first term of each A.P. is unity that means 1.
Let us assume the common difference for the A.P’s as ${{d}_{1}},{{d}_{2}},{{d}_{3}},......,{{d}_{n}}$ respectively and ${{n}^{th}}$ term for the A.P’s as ${{T}_{{{n}_{1}}}},{{T}_{{{n}_{2}}}},{{T}_{{{n}_{3}}}},......,{{T}_{{{n}_{n}}}}$. So, we have,
For A.P. 1, the terms can be written as: -
1, 2, 3, ……., n. Here, the ${{n}^{th}}$ term is n. So, ${{T}_{{{n}_{1}}}}=n$.
For A.P. 2, the terms can be written as: -
1, 3, 5, ……. upto n terms. So, applying the formula for ${{n}^{th}}$ term, we get,

& \Rightarrow {{T}_{{{n}_{2}}}}=a+\left( n-1 \right){{d}_{2}} \\\ & \Rightarrow {{T}_{{{n}_{2}}}}=1+\left( n-1 \right)\times 2 \\\ & \Rightarrow {{T}_{{{n}_{2}}}}=2n-1 \\\ \end{aligned}$$ For A.P. 3, the terms can be written as: - 1, 4, 7, 10, …. upto n terms. So, applying the formula for $${{n}^{th}}$$ term, we get, $$\begin{aligned} & \Rightarrow {{T}_{{{n}_{3}}}}=a+\left( n-1 \right){{d}_{3}} \\\ & \Rightarrow {{T}_{{{n}_{3}}}}=1+\left( n-1 \right)\times 3 \\\ & \Rightarrow {{T}_{{{n}_{3}}}}=3n-2 \\\ \end{aligned}$$ For A.P. 4, the terms can be written as: - 1, 5, 9, 13, ……… upto n terms. So, applying the formula for $${{n}^{th}}$$ term, we get, $$\begin{aligned} & \Rightarrow {{T}_{{{n}_{4}}}}=a+\left( n-1 \right){{d}_{4}} \\\ & \Rightarrow {{T}_{{{n}_{4}}}}=1+\left( n-1 \right)\times 4 \\\ & \Rightarrow {{T}_{{{n}_{4}}}}=4n-3 \\\ \end{aligned}$$ Similarly, on observing the pattern we can conclude the following expressions: - $$\Rightarrow {{T}_{{{n}_{5}}}}=5n-4,{{T}_{{{n}_{6}}}}=6n-5,.......,{{T}_{{{n}_{n}}}}=n.n-\left( n-1 \right)$$ Now, taking the sum of these $${{n}^{th}}$$ terms of all the A.P., we get, $$\Rightarrow S={{T}_{{{n}_{1}}}}+{{T}_{{{n}_{2}}}}+{{T}_{{{n}_{3}}}}+......+{{T}_{{{n}_{n}}}}$$ Here, S is the required sum. $$\Rightarrow S=n+\left( 2n-1 \right)+\left( 3n-2 \right)+\left( 4n-3 \right)+.....+\left( n.n-\left( n-1 \right) \right)$$ The above expression can be simplified as: - $$\begin{aligned} & \Rightarrow S=\left( n+2n+3n+4n+.....+n.n \right)-\left( 1+2+3+4+....+\left( n-1 \right) \right) \\\ & \Rightarrow S=n\left( 1+2+3+4+.....+n \right)-\left( 1+2+3+4+....+\left( n-1 \right) \right) \\\ \end{aligned}$$ Now, in the above expression, first term contains n terms in A.P. and second term contains (n - 1) terms in A.P., so applying the formula for sum of n terms and (n - 1) terms in an A. P, we get, $$\Rightarrow S=n\times {{S}_{n}}-{{S}_{n-1}}$$ Here, $${{S}_{n}}$$ and $${{S}_{n-1}}$$ denotes the sum of ‘n’ and (n - 1) terms respectively, so we have, $$\begin{aligned} & \Rightarrow S=n\times \dfrac{n}{2}\left[ 1+n \right]-\dfrac{n-1}{2}\left[ 1+\left( n-1 \right) \right] \\\ & \Rightarrow S=\dfrac{{{n}^{2}}}{2}\left( n+1 \right)-\dfrac{n-1}{2}\times n \\\ & \Rightarrow S=\dfrac{n}{2}\left[ n\left( n+1 \right)-\left( n-1 \right) \right] \\\ & \Rightarrow S=\dfrac{n}{2}\left[ {{n}^{2}}+n-n+1 \right] \\\ & \Rightarrow S=\dfrac{n}{2}\left[ {{n}^{2}}+1 \right] \\\ \end{aligned}$$ Hence proved **Note:** One may note that it is very important to form a general sequence for the $${{n}^{th}}$$ terms of all the A.P.’s. If we will not form any sequence then we will get confused in the terms and we will not be able to apply the formula for the sum of ‘n’ terms of an A.P. We have started all the A.P’s with 1 because it was given that the first term of all the A.P’s are 1.