Question

Find the sum upto n terms of the below series,
(1²) + (1²+2²) + (1²+2²+3²) + ...​

Answer 1

Hint: Write the general form of nth term of the given series in simplified form and then introduce summation to find the required sum upto nth term.

Solution:-

From the given series, we can observe that all the terms are sum of squares of n natural numbers.

Therefore, the general form of the given series is,

$\small{ \sf\implies T_k = {1}^{2} + {2}^{2} + {3}^{2} + ... + {k}^{2} }$

$\small{ \sf\implies T_k = \sum \limits_{k = 1}^k ({k}^{2}) }$

$\small{ \sf\implies T_k = \dfrac{k(k+ 1)(2k + 1)}{6} \qquad( \therefore formula \: used)}$

$\small{ \sf\implies T_k = \dfrac{( {k}^{2} + k)(2k + 1)}{6}}$

$\small{ \sf\implies T_k = \dfrac{2 {k}^{3} + 3 {k}^{2} + k }{6}}$

This is the general form of kth term of the given series.

Now, the sum of the given series is the sum of all the terms.

Therefore, the required sum is given by,

$\small{ \longrightarrow \sf\sum \limits_{k = 1}^n T_k}$

$\small{ \longrightarrow \sf\sum \limits_{k = 1}^n \left( \dfrac{2 {k}^{3} + 3 {k}^{2} + k }{6} \right) }$

$\small{ \longrightarrow \sf \frac{1}{6} \sum \limits_{k = 1}^n \left( 2 {k}^{3} + 3 {k}^{2} + k \right) }$

$\small{ \longrightarrow \sf \frac{1}{6} \left[\sum \limits_{k = 1}^n 2 {k}^{3} +\sum \limits_{k = 1}^n 3 {k}^{2} + \sum \limits_{k = 1}^nk\right] }$

$\small{ \longrightarrow \sf \frac{1}{6} \left[2\sum \limits_{k = 1}^n {k}^{3} +3\sum \limits_{k = 1}^n {k}^{2} + \sum \limits_{k = 1}^nk\right] }$

‎ ‎ ‎

We are aware about some formulas:-

• $\boxed{\small\tt\sum\limits_{k=1}^n k = \dfrac{n(n+1)}{2}}$

• $\boxed{\small\tt\sum\limits_{k=1}^n k^3 = \left[\dfrac{n(n+1)}{2}\right]^2}$

• $\boxed{\small\tt\sum\limits_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6}}$

‎ ‎ ‎

$\small{ \longrightarrow \sf \dfrac{1}{6} \left[2\left[\dfrac{n(n+1)}{2}\right]^2 +3. \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n + 1)}{2} \right] }$

$\small{ \longrightarrow \sf \dfrac{1}{6} \left[2\left[\dfrac{n(n+1)}{2}\right]^2 +3. \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n + 1)}{2} \right] }$

$\small{ \longrightarrow \sf \dfrac{1}{6} . \dfrac{n(n + 1)}{2} \left[2.\dfrac{n(n+1)}{2} +3. \dfrac{(2n+1)}{3} + 1\right] }$

$\small{ \longrightarrow \sf \dfrac{n(n + 1)}{12} \left[n(n+1) +(2n+1) + 1\right] }$

$\small{ \longrightarrow \sf \dfrac{n(n + 1)}{12} \left[ {n}^{2} + n +2n+1 + 1\right] }$

$\small{ \longrightarrow \sf \dfrac{n(n + 1)}{12} \left[ {n}^{2} +3 n + 2\right] }$

$\small{ \longrightarrow \sf \dfrac{n(n + 1)}{12} \left[ {n}^{2} + n+2n + 2\right] }$

$\small{ \longrightarrow \sf \dfrac{n(n + 1)}{12} \left[ n(n + 1) + 2(n + 1)\right] }$

$\small{ \longrightarrow \sf \dfrac{n(n + 1)}{12} \times (n + 2)(n + 1) }$

$\small{ \longrightarrow \sf \dfrac{n(n + 1)^{2}(n + 2) }{12} }$

This is the required sum upto nth term.