# Find the sum of the series 1⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + 4 ⋅ 5 +.....+ 8 ⋅ 9 ?

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Find the sum of the series 1⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + 4 ⋅ 5 +.....+ 8 ⋅ 9 ?
1. 240
2. 260
3. 225
4. 264
5. None of these

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Correct Answer - Option 1 : 240

Concept:

• Sum of the first n Natural Numbers $\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}$
• Sum of the Square of the first n Natural Numbers $\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}$
• Sum of the Cubes of the first n Natural Numbers $\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}$

Calculation:

Here, we have to find the sum of the series 1⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + 4 ⋅ 5 +.....+ 8 ⋅ 9

As we know that, the nth term of the given series is given by Tn = n (n + 1)

Sum of series = $\rm S_n = \sum T_n= \sum (n^2+n)=\sum n^2+\sum n$

As we know that, $\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}$

$⇒ \rm S_n = \frac{n(n+1)(n+2)}{3}$

Here, n = 8

$⇒ \rm S_8 = \frac{8 \;\times \;9 \;\times \;10}{3} = 240$

Hence, option 1 is the correct answer.