IIT JEE , AIEEE Forum - Mathematics , Algebra

Nithy

3 + 7 + 23 + 87...

I st diff is 4, 16, 64...(a GP)

so what should i do next???

6+ 9 + 16 + 27+....

I diff 3 7 11 ....(AP)

next????

help me!!! rates assured.

b_srikalyan009

3+7+23+87............

it can be written as

3+(3+4)+(3+4+4²)+(3+4+4²+4³).............................+[3+4(4^n-1- 1)/3 ]

hence

_{[1 ]}

^{[n ]} [ 3 + 4(4^r-1- 1)/3]

u can simplify now......

.........................................................................................................

similarly for the next question u can find the rth term and then find the sum.

the rth term is 6+(r-1)(2r-1)

while finding the summation u should remember sum(r) and sum(r²).

if u find this method useful please do rate me.

metal

Yes,

srikalyan's correct.

The first series can be written as:-

3+(3+4)+(3+4+4²)...........

The second one is simple also:-

It can be written as 6+(6+3)+(6+3+7)+(6+3+7+11).........(6+3+7+.......4n-5)

So, the nth term is 6+(n-1)(2n-1)=2n²-3n+7

Adding columnwise:-

t1=2(1²)-3(1)+7

t2=2(2²)-3(2)+7

.

..

tn=2(n²)-3(n)+7

adding we get:-

sumof n terms = 2(n)(n+1)(2n+1)/6 - 3(n)(n+1)/2+7n=(n(4n²-n+37))/6

By the way, if the sum is to be found till infinity, I must say that it is impossible

because the series does not converge. O.K?????????

please rate me ........

rajatsen91

good thinking metal and srikalyan

b_srikalyan009

if it is good then please rate

metal

I agree with srikalyan, if it is good then you should rate us , rajat

rohitsuv

Sum to n terms in simplified form for 3 + 7 + 23+ 87 to n terms
is 1/9(4^ (n+1) + 15n -4)

and for 6+9+16+27 + n terms is
n/6(4n^2 -3n +35)

metal

We already know that rohitsuv , don't we???????