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`        2+7+14+23+.... then find the sum upto 30 terms`
5 years ago

```							2, 7, 14, 23, 34 …........ is a series whose common difference is an AP (i.e   5, 7, 9, 11,….).S= 2 + 7 + 14 + 23 + ……………. An-1 + AnS=       2 +  7  + 14+ 23+………………+An-1  +AnNow, subtracting the equations0= 2 + { 5 + 7 +9 + 11 +…….+ (n-1)} - AnAn= 2+ { (n-1)(5+n-2)}On simplifying, we getAn = n2+2n-1Using the summation operator on the series we get∑An= ∑n2 + 2*∑n – ∑1 where,∑n2=n(n+1)(2n+1)/6∑n=n(n+1)/2 and∑1=n putting n=30 we get the required sum.;)
```
5 years ago
```							it is22-2   +   32-2    +    42-2     till                       312-2we will get is as (∑n2) -1*  – 2(30)                          -1* is taken because n2 starts from 12+22where∑n2=n(n+1)(2n+1)/6 on substituting the values it will be 30*31*61/6   = 9455so 9455-1-60=9394approve if useful
```
5 years ago
```							i think @ AUREA is it incorrect because AUREA forgot  one term in the  APPLIED AGPpl correct
```
5 years ago
```							22-2 = 232-2 =7 42-2 1452 -2=23 and so on approve if useful
```
5 years ago
```							 @akshat I think it should be S= ∑{(n+1)2-2}which simplifies to ∑(n2+2n-1)
```
5 years ago
```							putting the value n=30∑n2=94552∑n=930finally the sum comes to be S=9455+930-30 ,i.e, S=10355approve if the solution is helpful ; )
```
5 years ago
```							thanks aurea for correcting meinspite of n=31 i took it as n=30the answer would have been 10355 if had placed n=31sorry
```
5 years ago
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